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Electricity and Magnetism

Capacitance and Dielectrics

Capacitance is the capacity of a system of conductors to store electric charge per unit potential difference. It depends strictly on the geometrical configuration of the conductors and the dielectric medium separating them.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Calculation of capacitance for parallel plate, spherical, and cylindrical capacitors, energy stored, grouping of capacitors (series and parallel), and redistribution of charges.

PGT extension

Effect of dielectric polarization, displacement vector D, electric susceptibility, boundary conditions for electrostatic fields at dielectric interfaces, and transient response of RC circuits.

Prerequisites

Gauss's Law and its applications
Electric potential and potential energy
Boundary conditions of electrostatic fields

Definition and Physical Origin of Capacitance

Capacitance C is defined as the ratio of the magnitude of charge Q on either conductor to the potential difference V between them, expressed as $C = Q / V$.
The SI unit of capacitance is the farad (F), though practical values are typically in microfarads (uF) or picofarads (pF).
Capacitance is independent of the total charge Q and potential V; it is purely a geometric property determined by the shape, size, and separation of the conductors.

Dielectrics and Polarization

Insertion of a dielectric medium between capacitor plates reduces the electric field due to internal polarization, expressed as $E = E_0 / K$, where K is the dielectric constant.
The presence of a dielectric increases the capacitance by a factor of K, such that $C = K C_0$.
Bound surface charge density due to polarization is given by sigm$a_p = P · n = \sigma (1 - (1/K))$, where P is the polarization vector.

Energy Stored in a Capacitor

Electrostatic potential energy U stored in a charged capacitor is given by $U = (1/2) Q V = (1/2) C V^{2} = (1/2) (Q^{2} / C)$.
Energy density u (energy per unit volume) in an electrostatic field is given by $u = (1/2) \epsilon_0 E^{2} in free space$, and $u = (1/2) D · E in a dielectric medium$.
When two charged conductors are connected, redistribution of charge occurs, leading to a net loss of electrostatic energy, which is dissipated as heat and electromagnetic radiation.

Combination of Capacitors

In series combination, the charge Q on each capacitor is identical, and the reciprocal of equivalent capacitance is the sum of reciprocals: 1 / $C_s = \sum (1 / C_i)$.
In parallel combination, the potential difference V across each capacitor is identical, and the equivalent capacitance is the algebraic sum: $C_p = \sum C_i$.
Series combination reduces the equivalent capacitance to a value smaller than the smallest individual capacitance, while parallel combination increases it.

Derivations

Capacitance of a Parallel Plate Capacitor

Conditions: Two large conducting plates of area A separated by a small distance d in vacuum (d << sqrt(A)).

Apply Gauss's Law to find the electric field between the plates: $E = \sigma / \epsilon_0$, where sigm$a = Q / A$.
Determine the potential difference V between the plates by integrating the electric field: $V = E · d = (Q d) / (\epsilon_0 A)$.
Use the definition of capacitance $C = Q / V$ to substitute V.

Result: $C = \frac{\epsilon_0 A}{d}$

Energy Stored in a Parallel Plate Capacitor

Conditions: Charging a capacitor from zero charge to Q at a constant capacitance C.

Consider an instantaneous charge q on the capacitor plates and potential $v = q / C$.
The work required to transfer an infinitesimal charge dq is d$W = v dq = (q / C) dq$.
Integrate the work from q = 0 to final charge Q to obtain total work done: $U = \int_{0}^{Q} \frac{q}{C} dq$.

Result: $U = \frac{Q^2}{2C} = \frac{1}{2}CV^2$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Capacitance Definition	$$C = \frac{Q}{V}$$	C = Capacitance, Q = Charge, V = Potential difference F, C, V	Linear dielectric media
Parallel Plate Capacitor	$$C = \frac{\epsilon_0 A}{d}$$	A = Area of plates, d = Separation between plates, epsilo$n_0 = Permittivity of free space$ m², m, F/m	Uniform field approximation, edge effects neglected
Capacitor with Dielectric Slab	$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$	t = Thickness of dielectric slab, K = Dielectric constant m, dimensionless	Slab placed completely between plates
Spherical Capacitor	$$C = 4piepsilon_0 \frac{a b}{b - a}$$	a = Inner radius, b = Outer radius m	Concentric conducting spheres
Cylindrical Capacitor	$$C = \frac{2piepsilon_0 L}{\ln(b / a)}$$	L = Length of cylinders, a = Inner radius, b = Outer radius m	Coaxial cylinders, length L >> (b - a)
Series Combination	$$\frac{1}{C_s} = \sum_{i} \frac{1}{C_i}$$	$C_s = Equivalent series capacitance$ F	End-to-end connection
Parallel Combination	$$C_p = \sum_{i} C_i$$	$C_p = Equivalent parallel capacitance$ F	Connected across same potential nodes
Energy Density	$$u = \frac{1}{2}\epsilon_0 E^2$$	u = Energy density, E = Electric field intensity J/m³, V/m	Electrostatic field in vacuum

Graphs and experiments

Charge vs Potential Difference (Q-V Graph)

Draw: A straight line passing through the origin on a Q (y-axis) versus V (x-axis) coordinate system.

Interpret: The slope of the Q-V line represents the capacitance C of the device. The area under the Q-V curve represents the stored electrostatic energy U.

Exam use: Tested in conceptual questions regarding linear capacitors and energy calculation from graphical areas.

RC Circuit Charging Transient

Draw: An exponential growth curve for charge q(t) = Q_max (1 - e^(-t / RC)) versus time t.

Interpret: The time constant tau = RC characterizes the rate at which the capacitor charges, reaching approximately 63.2% of its maximum value at t = tau.

Exam use: Commonly featured in numericals determining time constants, asymptotic limits, and transient behaviors.

Capacitance vs Dielectric Insertion

Draw: A step or linear increase in capacitance as a dielectric slab of constant K is progressively inserted between the plates.

Interpret: Highlights the direct proportionality between capacitance and the dielectric constant K of the inserted medium.

Exam use: Evaluates understanding of polarization effects on stored charge and potential difference under isolated vs battery-connected conditions.

Solved numericals

TGT core

A parallel plate capacitor has circular plates each of radius 5 cm and a separation of 1 mm. Calculate its capacitance.

Identify given parameters: r = 0.05 m, d = 0.001 m, epsilo$n_0 = 8.85 x 10^-12 F/m$.
Calculate area of the circular plates: $A = \pi * r^2 = 3.1416 * (0.05)^2 = 7.854 x 10^-3 m^2$.
Substitute values into the parallel plate capacitance formula: $C = (\epsilon_0 * A) / d$.
$C = (8.85 x 10^-12 * 7.854 x 10^-3) / 10^-3 = 6.95 x 10^-11 F$.

Answer: 69.5 pF

PGT extension

A parallel plate capacitor of capacitance C is charged to potential V and then isolated. A dielectric slab of constant K = 4 and thickness equal to half the plate separation ($t = d/2)$ is inserted between the plates. Find the new potential difference.

Initial potential is $V = E_0 * d$, where $E_0 = \sigma / \epsilon_0$.
When the dielectric of thickness $t = d/2$ is inserted, the electric field inside the dielectric becomes $E = E_0 / K = E_0 / 4$, while in the remaining distance d/2, the field remains E_0.
The new potential difference V' is the sum of potential drops across the air gap and the dielectric: V' = E_0 * (d / 2) + (E_0 / 4) * (d / 2) = E_0 * d * (1/2 + 1/8) = V * (5/8).
Substitute V to get the final potential relation.

Answer: 5V / 8

PGT extension

Two capacitors of capacitances $C_1 = 2 uF$ and $C_2 = 3 uF$ are charged to potentials $V_1 = 100 V$ and $V_2 = 50 V respectively$, with plates of similar polarity connected together. Find the common potential and the loss of electrostatic energy.

Total initial charge $Q_total = C_1 * V_1 + C_2 * V_2 = (2 uF * 100 V) + (3 uF * 50 V) = 200 + 150 = 350 uC$.
Total equivalent capacitance in parallel $C_total = C_1 + C_2 = 2 uF + 3 uF = 5 uF$.
Common potential $V = Q_total / C_total = 350 uC / 5 uF = 70 V$.
Initial energy $U_i = (1/2) C_1 V_1^2 + (1/2) C_2 V_2^2 = 0.5 * 2 * 10^-6 * 10000 + 0.5 * 3 * 10^-6 * 2500 = 0.01 + 0.00375 = 0.01375 J$.
Final energy $U_f = (1/2) C_total V^2 = 0.5 * 5 * 10^-6 * (70)^2 = 0.01225 J$.
Energy loss delta $U = U_i - U_f = 0.01375 - 0.01225 = 0.00150 J$.

Answer: 70 V and 1.5 mJ

Assumptions and limitations

Edge effects at the periphery of parallel plates are neglected assuming infinite plate dimensions relative to separation (d << sqrt(A)).
Dielectric materials are assumed to be linear, homogeneous, and isotropic unless specified otherwise.

Common misconceptions

Common error: Capacitance depends on the total charge stored on the plates.

Correct: Capacitance is independent of charge and potential; it is determined solely by geometry and medium.

$C = Q / V$ is a constant ratio; increasing Q increases V proportionally, keeping C constant.

Common error: When a battery remains connected to a capacitor while a dielectric is inserted, the stored charge remains constant.

Correct: The potential difference remains constant while the charge increases by a factor of K.

A connected battery fixes the potential difference across the plates, forcing additional charge to flow from the source.

Quick revision

$C = Q/V = \epsilon_0 A / d$ for parallel plates.
Dielectric insertion multiplies capacitance and energy storage capacity by K.
Series combination decreases equivalent capacitance; parallel combination increases it.
Electrostatic energy loss occurs whenever two capacitors at different potentials are connected together.

Must remember

Capacitance $C = \frac{Q}{V}$ is a purely geometric property and is independent of the total charge $Q$ or potential difference $V$.
Insertion of a dielectric medium of dielectric constant $K$ increases capacitance to $C = K C_0$ and reduces the electric field to $E = \frac{E_0}{K}$ when charge is kept constant.
For a parallel plate capacitor, $C = \frac{\epsilon_0 A}{d}$. With a dielectric slab of thickness $t$, $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
Energy stored in a capacitor is given by $U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$.
Electrostatic energy density in free space is $u = \frac{1}{2}\epsilon_0 E^2$.
In series combination, $\frac{1}{C_s} = \sum \frac{1}{C_i}$; in parallel combination, $C_p = \sum C_i$.
When a battery remains connected during dielectric insertion, potential difference remains constant, and charge increases by factor $K$.
When a charged capacitor is disconnected and a dielectric is inserted, charge remains constant, and potential decreases by factor $K$.

Glossary

Capacitance ($C$)The ability of a system to store electric charge per unit potential difference, measured in Farads (F).

Dielectric Constant ($K$)A dimensionless quantity representing the ratio of the permittivity of a substance to the permittivity of free space.

Energy Density ($u$)The amount of electrostatic potential energy stored per unit volume in an electric field.

Polarization Vector ($\mathbf{P}$)The density of permanent or induced electric dipole moments in a dielectric material.

Bound Surface Charge Density ($\sigma_p$)The charge density appearing on the surface of a dielectric due to polarization under an electric field.

Series vs Parallel Combination of Capacitors

Parameter	Series Combination	Parallel Combination
Equivalent Capacitance	$\frac{1}{C_s} = \sum \frac{1}{C_i}$ (Smaller than smallest)	$C_p = \sum C_i$ (Larger than largest)
Charge ($Q$)	Same on each capacitor	Distributed as $Q_i = C_i V$
Potential Difference ($V$)	Divided as $V_i = \frac{Q}{C_i}$	Same across each capacitor

Practice Quiz

Choose an option to reveal the explanation.

1. What is the SI unit of capacitance?

Explanation: The SI unit of capacitance is the farad (F), defined as one coulomb per volt ($1 \text{ C/V}$).

2. The capacitance of a parallel plate capacitor depends upon:

Explanation: Capacitance is a purely geometric property determined by the area of the plates $A$, the separation $d$, and the intervening medium.

3. If a dielectric slab of dielectric constant $K$ is inserted completely between the plates of a charged isolated capacitor, the electric field between the plates becomes:

Explanation: Internal polarization of the dielectric reduces the electric field by a factor of $K$, so $E = \frac{E_0}{K}$.

4. What is the equivalent capacitance of three capacitors of values $2\ \mu\text{F}$, $3\ \mu\text{F}$, and $6\ \mu\text{F}$ connected in series?

Explanation: For series combination, $\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1\ \mu\text{F}^{-1}$, giving $C_s = 1\ \mu\text{F}$.

5. The energy stored in a capacitor of capacitance $C$ with a potential difference $V$ across its plates is given by:

Explanation: The electrostatic potential energy stored is $U = \frac{1}{2}CV^2$ (equivalently $\frac{1}{2}QV$ or $\frac{Q^2}{2C}$).

6. What happens to the energy stored in a charged capacitor when a dielectric slab is inserted while the battery remains connected?

Explanation: With the battery connected, $V$ is constant while $C$ becomes $KC_0$. The energy $U = \frac{1}{2}CV^2$ increases by a factor of $K$.

7. The formula for the capacitance of a spherical capacitor with inner radius $a$ and outer radius $b$ is:

Explanation: Integrating the electric field between concentric spheres of radii $a$ and $b$ yields $C = 4\pi\epsilon_0 \frac{ab}{b-a}$.

8. Electrostatic energy density $u$ in a vacuum electric field $E$ is expressed as:

Explanation: The energy stored per unit volume in an electric field in free space is $u = \frac{1}{2}\epsilon_0 E^2$.

9. Three capacitors of capacitances $2\ \mu\text{F}$, $3\ \mu\text{F}$, and $5\ \mu\text{F}$ are connected in parallel. Their equivalent capacitance is:

Explanation: In parallel, capacitances add algebraically: $C_p = 2 + 3 + 5 = 10\ \mu\text{F}$.

10. If the charge on a capacitor is doubled, its capacitance $C$ will:

Explanation: Capacitance is independent of charge and potential; doubling $Q$ causes $V$ to double, leaving $C = \frac{Q}{V}$ constant.

11. A parallel plate capacitor has plate area $A$ and separation $d$. If a metallic sheet of thickness $t$ ($t < d$) is inserted between the plates, the new capacitance is:

Explanation: For a conducting slab ($K \to \infty$), the effective separation becomes $d-t$, hence $C = \frac{\epsilon_0 A}{d-t}$.

12. What is the bound surface charge density $\sigma_p$ in terms of free surface charge density $\sigma$ and dielectric constant $K$?

Explanation: The polarization bound charge density is given by $\sigma_p = \sigma \left(1 - \frac{1}{K}\right)$.

13. When two charged capacitors at different potentials are connected in parallel:

Explanation: Charge is conserved, but energy is dissipated as heat and radiation during redistribution, resulting in a net loss of electrostatic energy.

14. The capacitance of a cylindrical capacitor of length $L$, inner radius $a$, and outer radius $b$ is given by:

Explanation: Integrating the electric field of coaxial cylinders yields $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$.

15. Which of the following describes the electric displacement vector $\mathbf{D}$?

Explanation: The electric displacement vector is defined as $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} = \epsilon \mathbf{E}$.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. A parallel plate air capacitor has a capacitance $C_0$. When a dielectric slab of thickness equal to half the plate separation and dielectric constant $K = 2$ is inserted, the new capacitance becomes:

Explanation: Using $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$. Here $t = \frac{d}{2}$ and $K = 2$. Thus $d - \frac{d}{2} + \frac{d/2}{2} = \frac{d}{2} + \frac{d}{4} = \frac{3d}{4}$. Therefore, $C = \frac{\epsilon_0 A}{3d/4} = \frac{4}{3} \left(\frac{\epsilon_0 A}{d}\right) = \frac{4}{3} C_0$.

2. Two identical capacitors each of capacitance $C$ and potential $V$ are connected in parallel with polarity reversed. The loss of electrostatic energy is:

Explanation: Initial energy $U_i = 2 \times \left(\frac{1}{2}CV^2\right) = CV^2$. When connected with reversed polarity, total charge is $Q_1 - Q_2 = 0$, final potential and energy are zero. Thus, loss of energy = $CV^2 - 0 = CV^2$.

3. A capacitor is charged by a battery and then disconnected. A dielectric slab is then inserted between the plates. Which of the following quantities remains constant?

Explanation: Since the capacitor is isolated (disconnected from the battery), the charge $Q$ on its plates cannot change.

4. The electric susceptibility $\chi_e$ of a dielectric medium is related to its dielectric constant $K$ by:

Explanation: The relationship between relative permittivity (dielectric constant $K$) and electric susceptibility $\chi_e$ is $K = 1 + \chi_e$, giving $\chi_e = K - 1$.

5. If $n$ identical capacitors, each of capacitance $C$, are connected first in parallel and then in series, the ratio of equivalent parallel capacitance to series capacitance ($C_p / C_s$) is:

Explanation: $C_p = nC$ and $C_s = \frac{C}{n}$. Therefore, $\frac{C_p}{C_s} = \frac{nC}{C/n} = n^2$.

6. A parallel plate capacitor with plate separation $d$ is charged to potential $V_0$. If a dielectric slab of thickness $d/2$ and $K=2$ is inserted, the potential difference across the plates (battery disconnected) becomes:

Explanation: Capacitance becomes $C = \frac{4}{3} C_0$. Since charge $Q$ is constant, $V = \frac{Q}{C} = \frac{Q}{(4/3)C_0} = \frac{3}{4} \left(\frac{Q}{C_0}\right) = \frac{3}{4} V_0$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the effective capacitance between terminals A and B if three capacitors of $3\ \mu\text{F}$ each are connected in the form of an equilateral triangle?

Explanation: Across any two terminals, one capacitor is in parallel with the series combination of the other two. Series pair = $\frac{3}{2} = 1.5\ \mu\text{F}$. Parallel with third = $1.5 + 3 = 4.5\ \mu\text{F}$.

2. The electric field in a dielectric medium is $E = 200\ \text{V/m}$ and the dielectric constant is $K = 5$. What is the polarization vector magnitude $P$ in $\text{C/m}^2$?

Explanation: Polarization $P = \epsilon_0 (K-1) E = \epsilon_0 (5-1)(200) = 4 \epsilon_0 (200)$.

3. Two conducting spheres of radii $R_1$ and $R_2$ are charged to the same potential and then connected by a thin conducting wire. The ratio of their final surface charge densities is:

Explanation: When connected, they attain equal potential $V_1 = V_2$. Surface charge density $\sigma = \frac{\epsilon_0 V}{R}$, so $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.

4. A capacitor of capacitance $C$ is charged to potential $V$ and connected across an uncharged capacitor of capacitance $2C$. The final common potential is:

Explanation: Conservation of charge: $CV = (C + 2C) V_f \implies 3C V_f = CV \implies V_f = \frac{V}{3}$.

5. If the electric field inside a dielectric slab is $E$ and the free charge density on plates is $\sigma$, the polarization $P$ is given by:

Explanation: From boundary conditions and Gauss's law in dielectrics, $E = \frac{\sigma - P}{\epsilon_0}$, which rearranges to $P = \sigma - \epsilon_0 E$.

6. In a parallel plate capacitor, the area of each plate is $A$ and separation is $d$. Half the space between plates is filled with a dielectric of constant $K_1$ and the other half with $K_2$ parallel to the plates. The equivalent capacitance is:

Explanation: When filled parallel to plates, it acts as two capacitors in parallel, each with area $A/2$ and separation $d$. $C = \frac{\epsilon_0 (A/2) K_1}{d} + \frac{\epsilon_0 (A/2) K_2}{d} = \frac{\epsilon_0 A}{2d}(K_1 + K_2)$.

7. The dimensional formula for capacitance $C$ is:

Explanation: Since $C = \frac{Q}{V} = \frac{It}{W/Q} = \frac{I^2 t^2}{ML^2 T^{-2}} = M^{-1} L^{-2} T^4 A^2$.

8. What is the energy stored per unit volume (energy density) in a dielectric medium where displacement is $D$ and electric field is $E$?

Explanation: In a dielectric medium, the electrostatic energy density is given by $u = \frac{1}{2} \mathbf{D} \cdot \mathbf{E}$.

9. A capacitor of $10\ \mu\text{F}$ is charged to $100\text{ V}$. Disconnected from the battery, it is connected in parallel to an uncharged capacitor. If the final potential is $40\text{ V}$, what is the capacitance of the second capacitor?

Explanation: Total charge $Q = 10\ \mu\text{F} \times 100\text{ V} = 1000\ \mu\text{C}$. Final voltage $40\text{ V} = \frac{1000}{10 + C_2} \implies 400 + 40C_2 = 1000 \implies 40C_2 = 600 \implies C_2 = 15\ \mu\text{F}$.

10. Boundary conditions at the interface of two dielectric media require which of the following components of the electric field $\mathbf{E}$ to be continuous across the boundary?

Explanation: The tangential component of the electric field ($\mathbf{E}_t$) is always continuous across the boundary between two media (i.e., $E_{1t} = E_{2t}$).