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Electrostatic Potential and Capacitance – CBSE NCERT Study Resources

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12th

12th - Physics

Electrostatic Potential and Capacitance

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Overview of the Chapter: Electrostatic Potential and Capacitance

This chapter introduces the concepts of electrostatic potential, potential difference, and capacitance, which are fundamental to understanding electric fields and their applications. Students will learn about equipotential surfaces, the relationship between electric field and potential, and the behavior of conductors and capacitors in electrostatic conditions.

Electrostatic Potential: The work done per unit charge to bring a small positive test charge from infinity to a point in an electric field without acceleration.

Key Topics Covered

  • Electric Potential and Potential Difference
  • Equipotential Surfaces
  • Potential Due to a Point Charge and Dipole
  • Capacitance and Capacitors
  • Combination of Capacitors
  • Energy Stored in a Capacitor
  • Dielectrics and Polarization

Electric Potential and Potential Difference

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric field. The potential difference between two points is the work done per unit charge in moving a charge between those points.

Potential Difference (V): The difference in electric potential between two points, measured in volts (V).

Equipotential Surfaces

An equipotential surface is a surface on which all points have the same electric potential. No work is required to move a charge along an equipotential surface, as the potential difference is zero.

Potential Due to a Point Charge and Dipole

The electric potential due to a point charge is given by \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge. For a dipole, the potential depends on the angle and distance relative to the dipole axis.

Capacitance and Capacitors

Capacitance is the ability of a conductor to store charge per unit potential difference. A capacitor is a device designed to store electric charge and energy in an electric field.

Capacitance (C): The ratio of the charge stored on a conductor to the potential difference across it, \( C = \frac{Q}{V} \).

Combination of Capacitors

Capacitors can be combined in series or parallel to achieve different capacitance values. In series, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances. In parallel, the equivalent capacitance is the sum of individual capacitances.

Energy Stored in a Capacitor

The energy stored in a capacitor is given by \( U = \frac{1}{2}CV^2 \), where \( C \) is the capacitance and \( V \) is the potential difference across the capacitor.

Dielectrics and Polarization

A dielectric is an insulating material that increases the capacitance of a capacitor when placed between its plates. Polarization of the dielectric reduces the effective electric field inside the capacitor.

Dielectric Constant (K): The ratio of the capacitance of a capacitor with a dielectric to the capacitance without the dielectric.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define electrostatic potential at a point.
Answer:

Work done to bring a unit positive charge from infinity to that point.

Question 2:
What is the SI unit of capacitance?
Answer:

Farad (F).

Question 3:
State the principle of a parallel plate capacitor.
Answer:

Stores energy in the electric field between two charged plates.

Question 4:
What happens to capacitance if the distance between plates of a capacitor increases?
Answer:

Capacitance decreases.

Question 5:
Name the factor affecting the dielectric strength of a material.
Answer:

Nature of the material.

Question 6:
What is the potential inside a charged spherical conductor?
Answer:

Constant and equal to surface potential.

Question 7:
Define equipotential surface.
Answer:

A surface where potential is the same at every point.

Question 8:
Give one example of a polar dielectric.
Answer:

Water (H2O).

Question 9:
How does a dielectric increase capacitance?
Answer:

Reduces effective electric field between plates.

Question 10:
What is the energy stored in a capacitor of capacitance C charged to potential V?
Answer:

(1/2)CV2.

Question 11:
State the effect of inserting a dielectric slab in a capacitor.
Answer:

Increases capacitance.

Question 12:
What is the potential gradient?
Answer:

Rate of change of potential with distance.

Question 13:
Why are equipotential surfaces perpendicular to electric field lines?
Answer:

No work is done along equipotential surfaces.

Question 14:
How does the potential energy of a system of two charges change if the distance between them is doubled?
Answer:

The potential energy (U) of a system of two charges is inversely proportional to the distance (r) between them.
If distance is doubled, U becomes half of its original value.
Mathematically, U ∝ 1/r.

Question 15:
What is the effect of a dielectric on the capacitance of a capacitor?
Answer:

Introducing a dielectric between the plates of a capacitor increases its capacitance.
The capacitance becomes K times the original value, where K is the dielectric constant of the material (K > 1).

Question 16:
What happens to the energy stored in a capacitor when it is connected to a battery and the plate separation is increased?
Answer:

When plate separation (d) is increased in a capacitor connected to a battery:
1. Capacitance decreases (C ∝ 1/d).
2. Since the battery maintains constant voltage, the energy stored (U = ½CV²) decreases.

Question 17:
Why is the electric field inside a conductor zero in electrostatic conditions?
Answer:

In electrostatic conditions, the electric field inside a conductor is zero because:
1. Free electrons redistribute themselves to cancel any external field.
2. Charges reside only on the surface, ensuring no net field inside.

Question 18:
What is the potential difference between two points in an electric field?
Answer:

The potential difference between two points is the work done per unit charge in moving a test charge from one point to another.
It is measured in volts (V) and given by ΔV = W/q.

Question 19:
How does the capacitance of a parallel plate capacitor change if the area of the plates is halved?
Answer:

The capacitance (C) of a parallel plate capacitor is directly proportional to the area (A) of the plates.
If area is halved, capacitance also becomes half (C ∝ A).

Question 20:
What is the significance of the dielectric strength of a material?
Answer:

The dielectric strength is the maximum electric field a dielectric can withstand without breakdown.
It determines the safe operating voltage for capacitors to prevent short-circuiting.

Question 21:
Why do capacitors block DC current after charging?
Answer:

Capacitors block DC current after charging because:
1. Once fully charged, no more charge flows through the capacitor.
2. DC has zero frequency, so the capacitive reactance (XC = 1/(2πfC)) becomes infinite, acting as an open circuit.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
How does the capacitance of a parallel plate capacitor change if the area of the plates is increased?
Answer:

The capacitance of a parallel plate capacitor is directly proportional to the area (A) of the plates.
If the area is increased, the capacitance increases proportionally.

Question 2:
Define electrostatic potential at a point. Write its SI unit.
Answer:

The electrostatic potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point against the electric field without any acceleration.
Its SI unit is volt (V).

Question 3:
What is the potential difference between two points in an electric field?
Answer:

The potential difference between two points in an electric field is the work done in moving a unit positive charge from one point to another against the electric field.
It is measured in volts (V).

Question 4:
State the relation between electric field and potential gradient.
Answer:

The electric field (E) is related to the potential gradient by the equation:
E = -dV/dr
where dV/dr is the potential gradient. The negative sign indicates that the electric field is directed from higher to lower potential.

Question 5:
What is an equipotential surface? Give one example.
Answer:

An equipotential surface is a surface on which the electric potential is the same at every point.
Example: The surface of a charged conductor in electrostatic equilibrium is an equipotential surface.

Question 6:
Define capacitance of a capacitor. Write its SI unit.
Answer:

The capacitance of a capacitor is defined as the ratio of the charge (Q) stored on either plate to the potential difference (V) between the plates.
C = Q/V
Its SI unit is farad (F).

Question 7:
What happens to the capacitance of a parallel plate capacitor if the distance between the plates is doubled?
Answer:

The capacitance of a parallel plate capacitor is inversely proportional to the distance (d) between the plates.
If the distance is doubled, the capacitance becomes half of its original value.

Question 8:
Why is the electric field inside a charged spherical conductor zero?
Answer:

Inside a charged spherical conductor, the electric field is zero because the charges redistribute themselves on the outer surface of the conductor, creating no net electric field inside.
This is a consequence of electrostatic shielding.

Question 9:
What is the effect of introducing a dielectric slab between the plates of a parallel plate capacitor?
Answer:

Introducing a dielectric slab between the plates of a parallel plate capacitor:

  • Increases the capacitance by a factor equal to the dielectric constant (K) of the material.
  • Reduces the electric field between the plates.

Question 10:
Derive the expression for energy stored in a capacitor.
Answer:

The energy (U) stored in a capacitor is given by:
U = (1/2)CV²
where C is the capacitance and V is the potential difference across the plates.
This energy is stored in the electric field between the plates.

Question 11:
What is the principle of a Van de Graaff generator?
Answer:

The Van de Graaff generator works on the principle of electrostatic induction and corona discharge.
It accumulates a large amount of charge on a hollow metal sphere, creating a high electrostatic potential.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define electrostatic potential at a point. Derive its expression for a point charge.
Answer:

The electrostatic potential at a point is defined as the work done per unit charge in bringing a small positive test charge from infinity to that point against the electric field without any acceleration.


For a point charge Q, the potential V at a distance r is derived as follows:


Work done to bring a test charge q from infinity to r is W = ∫∞r F·dr, where F = kQq/r2.


Thus, V = W/q = kQ/r, where k = 1/(4πε0).

Question 2:
Explain why the electric potential inside a charged conductor is constant.
Answer:

Inside a charged conductor, the electric potential is constant because:

  • The electric field inside a conductor in electrostatic equilibrium is zero (E = 0).
  • Since E = -dV/dr, if E = 0, then dV/dr = 0, meaning V does not change with position.

Thus, the conductor's surface acts as an equipotential surface, and the potential inside remains uniform.

Question 3:
Derive the expression for the capacitance of a parallel plate capacitor with a dielectric slab of thickness t between the plates.
Answer:

The capacitance of a parallel plate capacitor with a dielectric slab is derived as follows:


Let plate area = A, separation = d, and dielectric constant = K.


Effective separation reduces to d' = d - t + t/K due to the dielectric.


Capacitance C = ε0A/d' = ε0A/(d - t + t/K).


If the slab fills the gap (t = d), then C = Kε0A/d.

Question 4:
What is the significance of equipotential surfaces? Draw the equipotential surfaces for a point charge.
Answer:

Equipotential surfaces are surfaces where the potential is constant. Their significance includes:

  • No work is done to move a charge on an equipotential surface.
  • They are perpendicular to electric field lines.

For a point charge, equipotential surfaces are concentric spheres centered at the charge.


Diagram: A point charge with dotted concentric circles (spheres in 3D) around it, labeled as equipotential surfaces.

Question 5:
Explain how the energy stored in a capacitor changes when the separation between its plates is doubled while keeping the charge constant.
Answer:

The energy stored in a capacitor is given by U = Q2/(2C).


For a parallel plate capacitor, C = ε0A/d. If separation d is doubled, C becomes half.


Thus, U becomes Q2/(2(C/2)) = Q2/C, which is double the initial energy.


This is because work must be done to separate the plates against the attractive force between the charges.

Question 6:
Distinguish between conductors and dielectrics in terms of their behavior in an electric field.
Answer:

Conductors and dielectrics differ as follows:

  • Conductors: Free electrons move to neutralize the field inside, making E = 0. Charges reside on the surface.
  • Dielectrics: No free electrons; the field induces dipole alignment (polarization), reducing the net field but not nullifying it.

Conductors allow current flow, while dielectrics insulate and store energy.

Question 7:
Derive the expression for the capacitance of a parallel plate capacitor with a dielectric slab.
Answer:

For a parallel plate capacitor with plate area A, separation d, and dielectric constant K:


Without dielectric, capacitance C0 = ε0A/d.


With dielectric, the electric field reduces to E = E0/K, where E0 is the field without dielectric.


Thus, potential difference V = Ed = E0d/K.


Since Q = C0V0 = CV, substituting gives C = KC0 = Kε0A/d.

Question 8:
What is the significance of equipotential surfaces? Draw them for a dipole.
Answer:

Equipotential surfaces are surfaces where the potential is constant. Their significance includes:

  • No work is done to move a charge on such a surface.
  • They are perpendicular to electric field lines.

For a dipole, equipotential surfaces are curved and symmetric around the dipole axis. Near the charges, they are spherical, and farther away, they become asymptotic.


[Diagram: A dipole with curved equipotential surfaces and field lines perpendicular to them.]

Question 9:
Explain how energy is stored in a capacitor. Derive the expression for energy density.
Answer:

Energy is stored in a capacitor due to the separation of charges, creating an electric field between the plates.


The energy stored is U = ½CV2, where C is capacitance and V is potential difference.


For energy density (u), consider volume Ad between plates:


u = U/Ad = (½CV2)/Ad.


Substituting C = ε0A/d and V = Ed gives u = ½ε0E2.

Question 10:
Compare the series and parallel combinations of capacitors.
Answer:

Comparison:

  • Series: Total capacitance 1/Ceq = 1/C1 + 1/C2 + ....
    Charge is same on all capacitors.
    Voltage divides inversely with capacitance.
  • Parallel: Total capacitance Ceq = C1 + C2 + ....
    Voltage is same across all capacitors.
    Charge divides proportionally to capacitance.

Series reduces effective capacitance, while parallel increases it.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Derive the expression for the potential due to an electric dipole at a point on its axial line. Compare it with the potential on the equatorial line.
Answer:
Theoretical Framework

We studied that an electric dipole consists of two equal and opposite charges separated by a distance. The potential at a point due to a dipole depends on its position relative to the dipole axis.

Evidence Analysis
  • For axial position: V = (1/4πε₀) * (2p/r²), where p is dipole moment.
  • For equatorial position: V = 0, as the potentials due to +q and -q cancel out.
Critical Evaluation

The axial potential decreases as 1/r², while equatorial potential is always zero. This shows directional dependence of dipole fields.

Future Implications

This concept is crucial in understanding molecular interactions and dielectric behavior in capacitors.

Question 2:
Explain how the capacitance of a parallel plate capacitor changes when a dielectric slab is inserted between the plates. Derive the relevant expression.
Answer:
Theoretical Framework

Our textbook shows that capacitance C = ε₀A/d for parallel plates. When dielectric (k) is inserted, capacitance increases.

Evidence Analysis
  • With dielectric: C' = kε₀A/d
  • Derivation based on reduced effective field E = E₀/k
  • Example: Mica (k=6) increases capacitance sixfold
Critical Evaluation

This proves dielectrics allow higher charge storage at same potential, crucial for capacitor design.

Future Implications

Modern capacitors use high-k dielectrics for compact energy storage in electronics.

Question 3:
Describe the equipotential surfaces for (a) a single point charge (b) a uniform electric field. Explain why no work is done in moving a charge along such surfaces.
Answer:
Theoretical Framework

Equipotential surfaces have constant potential at all points. We studied their relation to electric fields.

Evidence Analysis
  • (a) For point charge: concentric spheres
  • (b) For uniform field: parallel planes
  • Work W = qΔV = 0 as ΔV=0
Critical Evaluation

This confirms electric field lines must be perpendicular to equipotential surfaces, following E = -dV/dr.

Future Implications

Understanding these surfaces helps in designing electrostatic shielding and particle accelerators.

Question 4:
Derive the expression for energy stored in a capacitor. Show that it can also be expressed in terms of electric field between the plates.
Answer:
Theoretical Framework

We know capacitors store energy when charged. The energy comes from work done in charging process.

Evidence Analysis
  • Basic form: U = ½CV²
  • Substituting C = ε₀A/d and V=Ed gives U = ½ε₀E²(Ad)
  • Example: 100μF at 50V stores 0.125J
Critical Evaluation

The field expression shows energy density depends on E², important for high-voltage applications.

Future Implications

This principle enables compact energy storage in modern supercapacitors.

Question 5:
Explain the principle of a Van de Graaff generator with a labeled diagram. How does it achieve very high potentials?
Answer:
Theoretical Framework

Van de Graaff generator uses electrostatic principles to build high voltage. Our textbook describes its working mechanism.

Evidence Analysis
  • Diagram: [Diagram: Belt, rollers, metal sphere]
  • Charge transfer via moving belt to hollow conductor
  • Potential V = Q/(4πε₀r) grows with accumulated charge
Critical Evaluation

The design prevents discharge until extremely high potentials (several million volts) are achieved.

Future Implications

Such generators remain vital for nuclear physics research and particle acceleration studies.

Question 6:
Derive the expression for the electric potential due to a point charge at a distance r from it. Explain the significance of the negative sign in the expression.
Answer:
Theoretical Framework

We studied that electric potential (V) at a point is the work done to bring a unit positive charge from infinity to that point. For a point charge q, the potential is derived using Coulomb's law and work-energy theorem.

Evidence Analysis
  • Using V = W/q, we integrate the electric field (E = kq/r²) over distance r.
  • The expression obtained is V = kq/r, where k is Coulomb's constant.
Critical Evaluation

The negative sign indicates that work is done against the electric field for a positive charge, aligning with conservative nature of electrostatic forces.

Question 7:
Compare the energy stored in two capacitors of the same capacitance, one connected to a battery and the other charged and disconnected. Justify your answer with derivations.
Answer:
Theoretical Framework

Energy stored in a capacitor (U) is given by U = ½CV². We compare two cases: capacitor connected to a battery (constant V) and isolated after charging (constant Q).

Evidence Analysis
  • Connected case: U1 = ½CV² remains constant as V is fixed.
  • Isolated case: If dielectric is inserted, C increases but Q is constant. Energy U2 = Q²/2C decreases.
Critical Evaluation

The battery supplies extra energy to maintain potential, while the isolated system conserves charge but loses energy due to work done in dielectric insertion.

Question 8:
Explain the principle of a parallel plate capacitor with a dielectric slab. Derive the expression for its capacitance and discuss the role of dielectric polarization.
Answer:
Theoretical Framework

A parallel plate capacitor stores energy by separating charges on two plates. Inserting a dielectric increases capacitance due to polarization.

Evidence Analysis
  • Capacitance without dielectric: C0 = ε0A/d.
  • With dielectric: C = Kε0A/d, where K is dielectric constant.
Critical Evaluation

Dielectric polarization reduces the net electric field, allowing more charge storage at the same potential. This is crucial in modern electronics for miniaturization.

Question 9:
Analyze the equipotential surfaces for a dipole. How do they differ from those of a single charge? Illustrate with a diagram.
Answer:
Theoretical Framework

Equipotential surfaces are 3D regions where potential is constant. For a dipole, they are non-concentric and asymmetric compared to a single charge's concentric spheres.

Evidence Analysis
  • Single charge: Spherical surfaces with V ∝ 1/r.
  • Dipole: Complex surfaces with V ∝ pcosθ/r² (where p is dipole moment).
[Diagram: Dipole equipotential surfaces showing peanut-shaped contours]Critical Evaluation

The asymmetry arises from the vector nature of dipole moments, influencing field orientation in applications like molecular physics.

Question 10:
Describe the Van de Graaff generator with a labeled diagram. Explain how the corona discharge limits its maximum potential.
Answer:
Theoretical Framework

The Van de Graaff generator uses a moving belt to accumulate high voltage on a hollow metal sphere through electrostatic induction.

[Diagram: Labeled components - belt, rollers, comb electrodes, sphere]Evidence Analysis
  • Charges are transferred via corona discharge at the sharp comb electrodes.
  • Maximum potential is limited when electric field at sphere's surface (~3×10⁶ V/m) ionizes surrounding air.
Critical Evaluation

This limitation demonstrates the practical constraints of electrostatic devices, though modern versions achieve ~7 MV using pressurized gas insulation.

Question 11:
Derive an expression for the energy stored in a parallel plate capacitor. Also, explain how this energy is related to the electric field between the plates.
Answer:

To derive the expression for the energy stored in a parallel plate capacitor, we start by considering the work done to charge the capacitor. Let C be the capacitance, V the potential difference, and Q the charge on the plates.


The work done (W) to move a small charge dq against the potential difference V is given by:


dW = V dq


Since V = q/C, we substitute this into the equation:


dW = (q/C) dq


To find the total work done to charge the capacitor from 0 to Q, we integrate:


W = ∫0Q (q/C) dq = (1/C) ∫0Q q dq


Evaluating the integral gives:


W = (1/C) [q2/2]0Q = Q2/2C


This work done is stored as electrostatic potential energy (U) in the capacitor. Thus:


U = Q2/2C


Using Q = CV, we can rewrite the energy as:


U = (CV)2/2C = (1/2)CV2


Alternatively, using C = ε0A/d (for a parallel plate capacitor with plate area A and separation d) and V = Ed (where E is the electric field), we get:


U = (1/2)(ε0A/d)(Ed)2 = (1/2)ε0E2Ad


Here, Ad is the volume between the plates. Thus, the energy density (u) (energy per unit volume) is:


u = U/Ad = (1/2)ε0E2


This shows that the energy stored in the capacitor is directly proportional to the square of the electric field between the plates.


Key Takeaways:

  • The energy stored in a capacitor is U = (1/2)CV2 or U = Q2/2C.
  • The energy density in the electric field is u = (1/2)ε0E2, highlighting the relationship between energy and the electric field.
Question 12:
Derive an expression for the electrostatic potential energy stored in a parallel plate capacitor. Also, explain how this energy is related to the electric field between the plates.
Answer:

The electrostatic potential energy (U) stored in a parallel plate capacitor can be derived as follows:


1. Consider a parallel plate capacitor with capacitance C, charge Q, and potential difference V between the plates.


2. The work done to charge the capacitor is stored as potential energy. The small work done (dW) to transfer a small charge dq against the potential V is:


dW = V dq = (q/C) dq


3. Total work done to charge the capacitor from 0 to Q is:


W = ∫0Q (q/C) dq = (1/C) ∫0Q q dq


4. Integrating gives:


W = (1/C) [q2/2]0Q = Q2/2C


5. This work is stored as potential energy (U):


U = Q2/2C = (1/2) CV2 = (1/2) QV


Relation to Electric Field:


The energy density (u) in the electric field (E) between the plates is given by:


u = (1/2) ε0 E2


where ε0 is the permittivity of free space. The total energy stored is the product of energy density and volume (Ad) between the plates:


U = u × Ad = (1/2) ε0 E2 × Ad


This shows that the energy is directly proportional to the square of the electric field.


Key Points:

  • The energy stored depends on the charge, capacitance, and voltage.
  • The electric field plays a crucial role in energy storage.
  • Energy density highlights the distribution of energy in the field.
Question 13:
Derive an expression for the electrostatic potential energy stored in a parallel plate capacitor. Explain how this energy is related to the electric field between the plates and the capacitance of the capacitor.
Answer:

The electrostatic potential energy (U) stored in a parallel plate capacitor can be derived as follows:


1. Consider a parallel plate capacitor with capacitance C, plate area A, and separation d. Let the charge on the plates be +Q and -Q, and the potential difference between them be V.


2. The work done to charge the capacitor is stored as potential energy. The energy stored when a small charge dq is transferred from one plate to the other against the potential difference V is:


dU = V dq


3. Since V = q/C (from the definition of capacitance), the total energy stored is obtained by integrating from 0 to Q:


U = ∫0Q (q/C) dq = (1/C) ∫0Q q dq


4. Solving the integral gives:


U = (1/C) [q²/2]0Q = Q²/2C


5. Using Q = CV, the energy can also be expressed as:


U = ½ CV² or U = ½ QV


Relation to the electric field and capacitance:

  • The energy density (energy per unit volume) in the electric field E between the plates is given by u = ½ ε₀E², where ε₀ is the permittivity of free space.
  • For a parallel plate capacitor, E = V/d and C = ε₀A/d. Substituting these into the energy expression shows how the energy depends on the field and geometry.

Thus, the energy stored is directly proportional to the square of the electric field and the capacitance of the system.

Question 14:
Define electric potential at a point. Derive an expression for the electric potential due to a point charge at a distance r from it. Explain how the potential varies with distance.
Answer:

The electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point against the electric field without any acceleration. It is a scalar quantity and is measured in volts (V).


Derivation for electric potential due to a point charge:


Consider a point charge Q placed at the origin. The electric field due to Q at a distance r is given by:


E = (1/4πε₀) * (Q/r²)


The work done in bringing a unit positive charge from infinity to a point at distance r is:


V = -∫∞r E · dr


Substituting E and solving the integral:


V = (1/4πε₀) * (Q/r)


Thus, the electric potential due to a point charge is inversely proportional to the distance r from the charge. As r increases, V decreases, following a 1/r relationship.


Variation with distance:


The potential is maximum near the charge and decreases as we move away from it. At infinity, the potential becomes zero.

Question 15:
Explain the principle of a parallel plate capacitor. Derive the expression for its capacitance when a dielectric slab of dielectric constant K is inserted between the plates.
Answer:

A parallel plate capacitor consists of two conducting plates separated by a small distance, with a dielectric medium (or air) in between. When charged, one plate acquires a positive charge, and the other an equal negative charge, creating a uniform electric field between them.


Derivation of capacitance with a dielectric slab:


Let the area of each plate be A, separation between plates be d, and the dielectric constant of the slab be K.


1. Without dielectric, capacitance is given by:


C₀ = ε₀A/d


2. When a dielectric is inserted, the electric field inside the dielectric reduces by a factor of K:


E = E₀/K


3. The potential difference between plates becomes:


V = E₀(d - t) + Et


where t is the thickness of the dielectric.


4. For a fully inserted slab (t = d):


V = E₀d/K


5. Since E₀ = σ/ε₀ and σ = Q/A, we get:


C = Q/V = (ε₀KA)/d


Thus, the capacitance increases by a factor of K when a dielectric is inserted.


Key points:

  • The dielectric reduces the effective electric field between the plates.
  • This allows more charge to be stored for the same potential difference.
  • The capacitance is directly proportional to the dielectric constant.
Question 16:
Derive an expression for the electrostatic potential energy stored in a capacitor. Explain how this energy is related to the work done in charging the capacitor.
Answer:

The electrostatic potential energy (U) stored in a capacitor is the energy required to charge it. Consider a capacitor of capacitance C charged to a potential difference V.


When a small charge dq is transferred from one plate to another, the work done (dW) is given by:


dW = V dq


Since V = q/C, the work done becomes:


dW = (q/C) dq


To find the total work done (W) to charge the capacitor from 0 to Q (final charge), we integrate:


W = ∫(q/C) dq from 0 to Q


This gives:


W = (1/2) (Q²/C)


Using Q = CV, the energy stored (U) can also be written as:


U = (1/2) CV² or U = (1/2) QV


This energy is equal to the work done in charging the capacitor. It is stored in the electric field between the plates.


Additional Insight: The energy density (energy per unit volume) in the electric field E is given by (1/2) ε₀E², showing how energy is distributed in the field.

Question 17:
Explain the concept of equipotential surfaces in an electric field. Draw the equipotential surfaces for (a) a single point charge and (b) a uniform electric field, and discuss their properties.
Answer:

Equipotential surfaces are imaginary surfaces where the electric potential is constant at every point. No work is done in moving a charge on such a surface because ΔV = 0.


Properties:

  • Work done to move a charge between two points on the same surface is zero.
  • Electric field lines are perpendicular to equipotential surfaces.
  • Two equipotential surfaces never intersect.
  • Closer surfaces indicate a stronger electric field.

(a) Single point charge:


For a positive point charge, equipotential surfaces are concentric spheres centered on the charge. The potential decreases radially outward.


Diagram: Nested spheres with the charge at the center, labeled with decreasing V values (V1 > V2 > V3).


(b) Uniform electric field:


In a uniform field (e.g., between parallel plates), equipotential surfaces are parallel planes perpendicular to the field lines. The potential decreases linearly in the direction of the field.


Diagram: Equally spaced parallel planes labeled with decreasing V values (V1 > V2 > V3), with uniform electric field lines perpendicular to them.


Key difference: The spacing between surfaces is uniform in a uniform field but varies (closer near the charge) for a point charge.

Question 18:
Derive an expression for the electrostatic potential energy stored in a system of two point charges. Explain how this energy varies with the distance between the charges.
Answer:

The electrostatic potential energy (U) of a system of two point charges is the work done in bringing the charges from infinity to their respective positions. Let us consider two point charges q1 and q2 separated by a distance r.


The potential energy of the system is given by:
U = k * (q1q2) / r
where k is Coulomb's constant (1/4πε0).


Derivation:
1. Initially, assume q1 is fixed at a point.
2. To bring q2 from infinity to a distance r from q1, work done (W) is stored as potential energy (U).
3. The work done is equal to the change in potential energy, which is:
W = U = q2 * V, where V is the potential due to q1 at the position of q2.
4. Since V = k * q1 / r, substituting gives:
U = k * (q1q2) / r.


Variation with distance:
1. The potential energy is inversely proportional to the distance r between the charges.
2. As r increases, U decreases, and vice versa.
3. For like charges (both positive or both negative), U is positive, indicating repulsive work is needed.
4. For unlike charges, U is negative, indicating attractive work is done by the system.

Question 19:
Explain the principle of a parallel plate capacitor and derive the expression for its capacitance. How does the capacitance change if a dielectric slab is inserted between the plates?
Answer:

A parallel plate capacitor consists of two conducting plates separated by a small distance, with a dielectric or air in between. It stores energy in the form of an electric field when a potential difference is applied.


Derivation of capacitance (C):
1. Let the area of each plate be A, separation be d, and the charge on the plates be +Q and -Q.
2. The electric field (E) between the plates is uniform and given by:
E = σ / ε0, where σ = Q / A (surface charge density).
3. The potential difference (V) between the plates is:
V = E * d = (Q / Aε0) * d.
4. Capacitance is defined as C = Q / V, so substituting V gives:
C = ε0A / d.


Effect of dielectric slab:
1. When a dielectric of permittivity ε (or dielectric constant K = ε / ε0) is inserted, the capacitance increases.
2. The new capacitance becomes:
C' = Kε0A / d.
3. The dielectric reduces the effective electric field between the plates, allowing more charge storage for the same potential difference.
4. The energy stored in the capacitor also increases if the voltage is kept constant.

Question 20:
Explain the principle of a parallel plate capacitor with a neat labeled diagram. Derive the expression for its capacitance when a dielectric slab of dielectric constant K is inserted between the plates.
Answer:

A parallel plate capacitor consists of two conducting plates separated by a small distance d, with area A each. When charged, one plate has charge +Q and the other -Q, creating an electric field between them.


Diagram (Description): Two parallel plates labeled +Q and -Q, separated by distance d, with an electric field E pointing from +Q to -Q.


Without dielectric, capacitance C₀ = ε₀A/d.


When a dielectric slab of constant K is inserted:


1. The electric field inside the dielectric reduces to E = E₀/K, where E₀ is the field without dielectric.


2. The potential difference V becomes V = E₀d/K.


3. Since Q = C₀V₀ = CV, substituting gives:


C = K C₀


Thus, the capacitance with dielectric is:


C = K ε₀A/d


Key Points:

  • The dielectric increases capacitance by a factor of K.
  • It reduces the electric field, preventing breakdown.
  • Practical applications include energy storage in devices.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A parallel-plate capacitor with air as the dielectric has a capacitance of 8 pF. The separation between the plates is doubled, and a dielectric slab of constant k = 4 is inserted. Case Deconstruction: How does this affect the capacitance? Theoretical Application: Calculate the new capacitance.
Answer:
Case Deconstruction:

Initially, capacitance (C) depends on plate area (A), distance (d), and permittivity (ε₀). Doubling 'd' reduces C by half. Inserting a dielectric (k=4) increases C by a factor of 'k'.


Theoretical Application:
  • Original C = 8 pF
  • After doubling 'd', C' = 8/2 = 4 pF
  • With dielectric, C'' = 4 × 4 = 16 pF
Question 2:
Two charges +5μC and -3μC are placed 10 cm apart in air. Case Deconstruction: What is the electrostatic potential energy of the system? Critical Evaluation: How does the energy change if the distance is halved?
Answer:
Case Deconstruction:

Potential energy (U) = kq₁q₂/r. Here, q₁ = +5μC, q₂ = -3μC, r = 0.1 m. U is negative as charges are opposite.


Critical Evaluation:
  • U = (9×10⁹)(5×10⁻⁶)(-3×10⁻⁶)/0.1 = -1.35 J
  • If r halves, U doubles (U ∝ 1/r). New U = -2.7 J
Question 3:
A 12V battery is connected to a 6μF capacitor. Case Deconstruction: What charge is stored? The battery is disconnected, and a dielectric (k=3) is inserted. Theoretical Application: Find the new potential difference.
Answer:
Case Deconstruction:

Charge (Q) = CV = 6μF × 12V = 72μC. After disconnection, Q remains constant. Inserting dielectric increases C by 'k'.


Theoretical Application:
  • New C' = 6μF × 3 = 18μF
  • V' = Q/C' = 72μC/18μF = 4V
Question 4:
A spherical conductor of radius 12 cm has a charge of 1.6×10⁻⁷C. Case Deconstruction: Determine its capacitance. Critical Evaluation: How does capacitance change if the charge is doubled?
Answer:
Case Deconstruction:

Capacitance (C) = 4πε₀r. For r = 0.12 m, C = 0.12/(9×10⁹) ≈ 13.3 pF.


Critical Evaluation:
  • C depends only on geometry (radius), not charge.
  • Doubling charge doesn’t alter C (still ~13.3 pF).
Question 5:
Two capacitors (3μF and 6μF) are connected in series to a 24V battery. Case Deconstruction: Calculate the equivalent capacitance. Theoretical Application: Find the charge on each capacitor.
Answer:
Case Deconstruction:

In series, 1/Ceq = 1/C₁ + 1/C₂. Here, C₁ = 3μF, C₂ = 6μF.


Theoretical Application:
  • Ceq = (3×6)/(3+6) = 2μF
  • Charge (Q) = CeqV = 2μF × 24V = 48μC (same for both).
Question 6:
A parallel plate capacitor with air as the dielectric has a capacitance of 8 pF. The separation between the plates is doubled, and a dielectric slab of dielectric constant 6 is inserted. Calculate the new capacitance. Justify your steps using the formula for capacitance.
Answer:
Case Deconstruction

We know capacitance (C) = ε₀A/d. Initially, C₁ = 8 pF. When plate separation (d) is doubled, C₂ = C₁/2 = 4 pF (since C ∝ 1/d).

Theoretical Application

After inserting the dielectric (K = 6), C₃ = KC₂ = 6 × 4 pF = 24 pF. Our textbook shows that capacitance increases by a factor of K when a dielectric is introduced.

Critical Evaluation

The calculation aligns with the formula C = Kε₀A/d. Doubling d reduces C by half, but the dielectric compensates by increasing it sixfold.

Question 7:
Two identical capacitors are connected in series and charged with a 12V battery. If one capacitor is removed, how does the potential difference across the remaining capacitor change? Explain using charge conservation.
Answer:
Case Deconstruction

In series, total capacitance Cₜ = C/2. Charge Q = CₜV = (C/2) × 12V = 6C µC. Each capacitor stores this charge.

Theoretical Application

When one capacitor is removed, the remaining capacitor has the same charge (6C µC) but capacitance C. Thus, V = Q/C = 6V. Our textbook confirms charge remains conserved in isolated systems.

Critical Evaluation

The potential difference halves because the same charge is now stored in a single capacitor with double the original effective capacitance.

Question 8:
A 10 µF capacitor is charged to 100V and disconnected. It is then connected to an uncharged 20 µF capacitor. Calculate the common potential and energy loss. Validate with the energy conservation principle.
Answer:
Case Deconstruction

Initial charge Q = C₁V₁ = 10 µF × 100V = 1000 µC. When connected, total capacitance Cₜ = 10 + 20 = 30 µF.

Theoretical Application

Common potential V = Q/Cₜ = 1000 µC / 30 µF ≈ 33.33V. Energy loss = ½C₁V₁² - ½CₜV² ≈ 5 mJ - 1.67 mJ = 3.33 mJ. Our textbook shows energy is lost as heat during redistribution.

Critical Evaluation

The loss aligns with the principle that energy is conserved only in ideal systems without resistance.

Question 9:
Derive the expression for equivalent capacitance of three capacitors C₁, C₂, and C₃ connected in parallel. Compare it with their series combination using an example where C₁ = 2 µF, C₂ = 3 µF, and C₃ = 6 µF.
Answer:
Case Deconstruction

In parallel, Cₜ = C₁ + C₂ + C₃. For the given values, Cₜ = 2 + 3 + 6 = 11 µF. In series, 1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃.

Theoretical Application

For series, 1/Cₜ = ½ + ⅓ + ⅙ = 1 ⇒ Cₜ = 1 µF. Our textbook shows parallel combination increases Cₜ, while series reduces it.

Critical Evaluation

The example confirms parallel Cₜ is always greater than individual capacitances, whereas series Cₜ is smaller than the smallest capacitor.

Question 10:
A spherical conductor of radius 12 cm has a charge of 1.6 µC. Calculate the electrostatic potential at its surface and at a point 10 cm from its center. Justify the difference using Gauss’s law.
Answer:
Case Deconstruction

Potential at surface V = kQ/r = (9 × 10⁹ × 1.6 × 10⁻⁶)/0.12 = 120 kV. Inside the conductor (10 cm < 12 cm), V is constant (120 kV).

Theoretical Application

Gauss’s law states E = 0 inside a conductor, so potential remains uniform. Our textbook confirms potential inside equals surface potential for conductors.

Critical Evaluation

The result aligns with the property of conductors being equipotential surfaces, as no work is done to move charge inside.

Question 11:
A parallel-plate capacitor with air as the dielectric has a capacitance of 8 pF. The separation between the plates is doubled, and a dielectric slab of dielectric constant 6 is inserted. Calculate the new capacitance. Justify your answer using the formula for capacitance with a dielectric.
Answer:
Case Deconstruction

Initial capacitance (C0) = 8 pF. Separation (d) is doubled, and a dielectric (K=6) is inserted.


Theoretical Application

We studied that capacitance (C) = (Kε0A)/d. When d doubles, C halves (4 pF). With dielectric, C becomes K×Cnew = 6×4 pF = 24 pF.


Critical Evaluation

Our textbook shows that inserting a dielectric increases capacitance proportionally to K. Here, the effect of increased d is offset by the dielectric.

Question 12:
Two identical capacitors are connected in series and charged to a potential difference of 120 V. If one capacitor is removed, how does the energy stored in the system change? Derive the ratio of final to initial energy.
Answer:
Case Deconstruction

Series connection: Ceq = C/2. Energy (U) = ½CeqV². One capacitor removed: Ceq = C.


Theoretical Application

Initial Ui = ½(C/2)(120)² = 3600C. Final Uf = ½C(120)² = 7200C. Ratio Uf/Ui = 2:1.


Critical Evaluation

We studied that energy in series is less than individual capacitors. Removing one doubles the energy, as stored energy ∝ 1/Ceq.

Question 13:
A 10 μF capacitor is charged to 50 V and disconnected. A second uncharged 20 μF capacitor is connected parallel to it. Calculate the common potential and energy loss. Explain the loss mechanism.
Answer:
Case Deconstruction

Initial charge (Q) = 10 μF × 50 V = 500 μC. Parallel connection: Ctotal = 30 μF.


Theoretical Application

Common potential (V) = Q/Ctotal = 500/30 = 16.67 V. Energy loss = ½(10)(50)² − ½(30)(16.67)² = 4166.67 μJ.


Critical Evaluation

Our textbook shows energy loss occurs due to charge redistribution, often as heat. This aligns with the principle of conservation of charge.

Question 14:
An isolated conducting sphere of radius 10 cm has a capacitance of 10 pF. If its radius is halved, how does its capacitance change? Relate this to the formula for spherical capacitors.
Answer:
Case Deconstruction

Initial radius (R) = 10 cm, C = 10 pF. New radius = 5 cm.


Theoretical Application

We studied C = 4πε0R for spheres. C ∝ R. Halving R halves C. New C = 5 pF.


Critical Evaluation

This matches NCERT’s derivation: capacitance depends linearly on radius for isolated conductors. Smaller radius reduces charge storage capacity.

Question 15:
A 4 μF capacitor is charged to 100 V and connected to an uncharged 2 μF capacitor. Calculate the potential difference across each capacitor post-connection and the charge redistribution.
Answer:
Case Deconstruction

Initial charge (Q) = 4 μF × 100 V = 400 μC. Total capacitance post-connection = 6 μF.


Theoretical Application

Common potential (V) = Q/Ctotal = 400/6 = 66.67 V. Charges: Q4μF = 4×66.67 = 266.68 μC, Q2μF = 2×66.67 = 133.33 μC.


Critical Evaluation

Our textbook shows charge redistributes to equalize potential. The example confirms charge conservation (266.68 + 133.33 ≈ 400 μC).

Question 16:

A parallel plate capacitor is charged by a battery and then disconnected. The plates are then slowly pulled apart to double the initial separation. Based on this scenario, answer the following:

  • How does the capacitance of the capacitor change? Justify your answer.
  • What happens to the energy stored in the capacitor? Explain with reasoning.
Answer:

Capacitance (C) of a parallel plate capacitor is given by the formula: C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.


When the separation is doubled (d' = 2d), the capacitance becomes: C' = ε₀A/(2d) = C/2. Thus, the capacitance halves.


The energy stored (U) in a capacitor is given by: U = Q²/2C, where Q is the charge (constant here as the battery is disconnected).


Since C decreases to C/2, the energy becomes: U' = Q²/(2 × C/2) = Q²/C = 2U. Hence, the energy stored doubles.


This happens because work is done to pull the plates apart against the electrostatic attraction, increasing the system's potential energy.

Question 17:

Two identical capacitors, C₁ and C₂, are connected in series and charged by a battery of voltage V. After disconnecting the battery, a dielectric slab of dielectric constant K is inserted into C₁. Analyze the following:

  • How does the potential difference across each capacitor change? Provide a detailed explanation.
  • What happens to the total energy of the system? Derive the expression for the new energy.
Answer:

Potential Difference Change: Initially, both capacitors have equal capacitance C, so the potential difference across each is V/2 (since they are in series).
After inserting the dielectric, the capacitance of C₁ becomes C₁' = KC, while C₂ remains C.
The charge Q remains constant (battery disconnected), so the new potential differences are:
V₁' = Q/(KC) and V₂' = Q/C.
Since V₁' + V₂' = V, solving gives V₁' = V/(K + 1) and V₂' = KV/(K + 1).
Thus, V₁' decreases, while V₂' increases due to the redistribution of charge.

Total Energy: Initial energy is Uinitial = ½CeqV² = ½(C/2)V² = CV²/4.
Final energy is Ufinal = ½(Q²/(KC) + Q²/C), where Q = CV/2.
Substituting Q, we get Ufinal = (CV²/8)(1/K + 1).
The energy decreases because the dielectric reduces the electric field inside C₁, lowering the system's energy.

Question 18:

Two identical capacitors, C₁ and C₂, are connected in series to a battery of voltage V. After fully charging, a dielectric slab of dielectric constant K is inserted into C₁. Analyze the following:

  • How does the potential difference across each capacitor change? Explain.
  • What happens to the total energy stored in the system? Provide a detailed justification.
Answer:

Initially, both capacitors have equal capacitance C, so in series, the equivalent capacitance is Ceq = C/2, and the charge on each is Q = (C/2)V.


After inserting the dielectric into C₁, its new capacitance becomes C₁' = KC, while C₂ remains C.


The new equivalent capacitance is: Ceq' = (KC × C)/(KC + C) = KC/(K + 1).


The charge redistributes as Q' = Ceq'V = (KC/(K + 1))V.


The potential difference across C₁' is: V₁' = Q'/C₁' = (KC/(K + 1))V / (KC) = V/(K + 1).


Across C₂, it is: V₂' = Q'/C₂ = (KC/(K + 1))V / C = KV/(K + 1).


Thus, V₁' decreases while V₂' increases.


The total energy stored initially was: U = (1/2)(C/2)V² = CV²/4.


After insertion, it becomes: U' = (1/2)Ceq'V² = (1/2)(KC/(K + 1))V².


Comparing U and U', U' < U because K/(K + 1) < 1. The energy decreases due to the dielectric reducing the electric field inside C₁.

Question 19:
A parallel plate capacitor is charged by a battery and then disconnected. The plates are then slowly pulled apart. Analyze the effect on:
(i) Capacitance
(ii) Potential difference between plates
(iii) Energy stored in the capacitor.
Answer:

When a parallel plate capacitor is disconnected from the battery and the plates are pulled apart:


(i) Capacitance (C): The capacitance decreases because C = ε₀A/d, where A is the plate area and d is the separation. As d increases, C decreases.

(ii) Potential difference (V): Since the charge Q remains constant (battery disconnected), and V = Q/C, the potential difference increases as C decreases.

(iii) Energy stored (U): The energy stored is given by U = Q²/2C. As C decreases, U increases because work is done to separate the plates against the electrostatic attraction.
Question 20:
Two identical capacitors are connected first in series and then in parallel across the same battery. Compare the total energy stored in both configurations.
Answer:

Let each capacitor have capacitance C and the battery voltage be V.


Case 1: Series Connection
Equivalent capacitance: Ceq = C/2
Energy stored: Useries = ½ (C/2) V² = CV²/4

Case 2: Parallel Connection
Equivalent capacitance: Ceq = 2C
Energy stored: Uparallel = ½ (2C) V² = CV²

Comparison: The parallel configuration stores 4 times more energy than the series configuration because the equivalent capacitance is higher, allowing more charge storage at the same voltage.
Question 21:
A parallel plate capacitor is charged by a battery and then disconnected. The plates are then slowly pulled apart. Explain the effect on:
(i) The capacitance of the capacitor
(ii) The potential difference between the plates
(iii) The energy stored in the capacitor
Answer:

When the plates of a parallel plate capacitor are pulled apart after disconnecting the battery:


(i) Capacitance (C): The capacitance decreases because C = ε₀A/d, where d (distance between plates) increases while A (area) and ε₀ remain constant.

(ii) Potential Difference (V): Since the charge Q remains constant (battery disconnected), V = Q/C increases as C decreases.

(iii) Energy Stored (U): The energy U = Q²/2C increases because C decreases while Q is constant. Work is done to separate the plates against the electrostatic attraction, converting mechanical energy into electrical potential energy.
Question 22:
Two identical metal spheres, A and B, are placed in contact. A positively charged rod is brought near sphere A without touching it. While the rod is held close, the spheres are separated. Describe the final charge distribution on spheres A and B and justify your answer.
Answer:

When the positively charged rod is brought near sphere A:


Step 1: Due to electrostatic induction, negative charges in the spheres migrate toward sphere A (closer to the rod), while positive charges are repelled to sphere B.

Step 2: While the rod is still near, separating the spheres traps the charges: sphere A acquires a net negative charge, and sphere B acquires an equal positive charge.

Justification: The rod induces a charge redistribution without direct contact. The total charge remains zero (conservation of charge), but the separation fixes the induced charges on each sphere.
Question 23:
A parallel plate capacitor is charged by a battery and then disconnected. The plates are then slowly pulled apart. Explain the changes in the following quantities: (i) Capacitance, (ii) Charge, (iii) Potential difference, and (iv) Energy stored in the capacitor.
Answer:

When the plates of a charged capacitor are pulled apart after disconnecting the battery:


(i) Capacitance (C): The capacitance decreases because C = ε₀A/d, where d (distance between plates) increases while A (area) and ε₀ remain constant.

(ii) Charge (Q): The charge remains unchanged because the capacitor is disconnected, and there is no path for charge to flow.

(iii) Potential difference (V): The potential difference increases as V = Q/C. Since Q is constant and C decreases, V must increase.

(iv) Energy stored (U): The energy stored increases because U = Q²/2C. With Q constant and C decreasing, U increases. The work done in pulling the plates apart converts into the increased energy of the capacitor.

Question 24:
Two identical metal spheres, A and B, are placed in contact. A positively charged rod is brought near sphere A without touching it. The spheres are then separated, and the rod is removed. Explain the final charge distribution on spheres A and B.
Answer:

When a positively charged rod is brought near sphere A:


Step 1: Due to electrostatic induction, negative charges in sphere A are attracted toward the rod, while positive charges are repelled to sphere B.

Step 2: While the rod is still near, the spheres are separated. Sphere A retains the induced negative charge, and sphere B retains the positive charge.

Step 3: When the rod is removed, the charges redistribute slightly, but the net charge remains: sphere A is negatively charged, and sphere B is positively charged.

This happens because the rod induces a charge separation, and separating the spheres while the rod is present 'locks' the charges in place.

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