Capacitors And Capacitance

Q: What is the difference between a conductor's capacitance and a capacitor's capacitance?

A single isolated conductor has capacitance C = Q/V where V is measured from infinity. A capacitor has two conductors, and its capacitance C = Q/ΔV uses the potential difference between the two plates. Capacitor values are much larger than isolated-conductor values for the same size.

Q: Does capacitance depend on the material of the plates?

No. Capacitance depends only on the geometry (size, shape, plate separation) and on the dielectric medium between the plates, not on which specific metal is used.

Q: Why is a farad such a large unit that we rarely see capacitors rated in whole farads?

One farad means storing 1 coulomb at just 1 volt. To get 1 F from a parallel plate capacitor with a 1 mm gap, you would need plates about 110 km squared in area. Everyday electronics use microfarad, nanofarad or picofarad values.

Q: In a series combination of capacitors, is the middle capacitor's charge really the same as the end ones?

Yes. In a series chain the same current flows through the entire branch during charging. Since capacitor charge is the time integral of current, each capacitor accumulates the same charge.

PhysicsCapacitorsFor NEET aspirants

A capacitor is a device that stores electric charge and energy in the electric field between two conductors separated by an insulator. Its capacitance is the ratio of stored charge to the resulting potential difference, $C = Q / V$ , measured in farads. Understanding capacitance and how capacitors combine in series and parallel is fundamental for JEE and NEET Physics electrostatics problems, particularly circuit analysis and energy storage questions.

Key Formulas - Quick Reference

Capacitance definition: $C = Q / V$ (unit: farad, F)
Isolated spherical conductor: $C = 4 π ε_{0} R$
Isolated sphere in medium of dielectric constant $K$ : $C_{m} = 4 π ε_{0} K R$
Common potential when two capacitors connect: $V = (C_{1} V_{1} + C_{2} V_{2}) / (C_{1} + C_{2})$
Energy lost during charge sharing: $Δ U = \frac{1}{2} \cdot \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$
Series: $1/ C_{e q} = 1/ C_{1} + 1/ C_{2} + 1/ C_{3} + \dots$ ; charge is same on each
Parallel: $C_{e q} = C_{1} + C_{2} + C_{3} + \dots$ ; voltage is same across each

1. What is Capacitance?

When a conductor is given a charge $Q$ , its potential $V$ rises in proportion to that charge:

Q \propto V \Rightarrow Q = C V \Rightarrow C = \frac{Q}{V}

The proportionality constant

C

is the capacitance of the conductor. It is a measure of the conductor's ability to store electric charge for a given potential.

Key properties

Capacitance is a measure of a conductor's ability to hold electric charge.
Every conductor has a charging limit. Beyond that, the surrounding medium ionizes and charge leaks away.
Capacitance $C$ depends on: (i) size and shape of the conductor, (ii) surrounding medium, (iii) presence of other conductors nearby.
Capacitance does not depend on the amount of charge given or the material of the conductor.

Units and dimensions

SI unit: farad (F). $1 F = 1 C / V$ . A farad is very large; practical values use $μ F$ , $n F$ , $p F$ .
CGS unit: statfarad. $1 F = 9 \times 1 0^{11} statfarad$ .
Dimensions: $[C] = [M^{- 1} L^{- 2} T^{4} A^{2}]$ .

2. Capacitance of an Isolated Spherical Conductor

Figure: Isolated spherical conductor. Capacitance C = 4πε₀R depends only on radius R.

For an isolated conducting sphere of radius $R$ carrying charge $Q$ , the potential at its surface is:

V = \frac{Q}{4 π ε _{0} R} and V = \frac{Q}{C}

\Rightarrow C = 4 π ε_{0} R

Some direct consequences:

Capacitance depends only on the radius, not on the charge given.
If the surrounding medium has dielectric constant $K$ , capacitance becomes $C_{m} = 4 π ε_{0} K R$ .
$C \propto R$ and $C \propto K$ .
Since $ε_{0} = C / (4 π R)$ , the units of $ε_{0}$ are $F / m$ , which is the same as $C^{2} / (N \cdot m^{2})$ .

3. Charge Distribution Between Two Connected Conductors

Take two insulated spherical conductors $A$ and $B$ with capacitances $C_{1}$ and $C_{2}$ , carrying charges $q_{1}$ and $q_{2}$ at potentials $V_{1}$ and $V_{2}$ respectively. So $q_{1} = C_{1} V_{1}$ and $q_{2} = C_{2} V_{2}$ .

When they are connected by a thin wire, positive charge flows from higher to lower potential until both reach a common potential $V$ . Charge is conserved throughout:

V = \frac{Total charge}{Total capacitance} = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}

Distributed charges after sharing:

q_{1}^{'} = C_{1} V, q_{2}^{'} = C_{2} V \Rightarrow \frac{q _{1}^{'}}{q _{2}^{'}} = \frac{C _{1}}{C _{2}}

Charge distributes in the ratio of capacitances, not in the ratio of initial charges.

Energy loss during sharing

The total electrostatic energy stored decreases because heat is dissipated in the connecting wire (and some energy is radiated):

Δ U = U_{initial} - U_{final} = \frac{1}{2} \cdot \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}

Even with a superconducting wire (zero resistance), the two capacitors still cannot end up with the same energy they started with. The lost energy escapes as electromagnetic radiation. This is a genuine physical loss, not a mathematical artifact.

4. Capacitor (Condenser)

Since the capacitance of a single conductor depends on nearby conductors, we can boost it deliberately by placing another conductor close to the first. This arrangement of two conductors, carrying equal and opposite charges, is a capacitor (also called a condenser in older texts).

For a capacitor with charge

Q

on one plate,

- Q

on the other, and potential difference

Δ V

across them:

C = \frac{Q}{Δ V}

5. Combination of Capacitors

Capacitors combine in two basic ways: series and parallel. Any complex network reduces to these building blocks.

Series combination

Figure: Three capacitors in series. Charge Q is the same on each; voltages V1, V2, V3 add to V.

Charge is the same on every capacitor: $Q = C_{1} V_{1} = C_{2} V_{2} = C_{3} V_{3} = \dots$
The potential across each capacitor is inversely proportional to its capacitance: $V \propto 1/ C$ .
Total potential difference across the combination: $V = V_{1} + V_{2} + V_{3} + \dots$
Equivalent capacitance formula: $\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} + \dots$

For just two capacitors in series:

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}, V_{1} = \frac{C _{2}}{C _{1} + C _{2}} V, V_{2} = \frac{C _{1}}{C _{1} + C _{2}} V

Parallel combination

Figure: Three capacitors in parallel. Same voltage V across each; charges Q1, Q2, Q3 add to total Q.

Potential difference is the same across every capacitor, equal to the applied voltage $V$ .
Charge on each capacitor is proportional to its capacitance: $Q_{1} = C_{1} V, Q_{2} = C_{2} V, Q_{3} = C_{3} V, \dots$
Total charge from the source: $Q = Q_{1} + Q_{2} + Q_{3} + \dots$
Equivalent capacitance: $C_{e q} = C_{1} + C_{2} + C_{3} + \dots$

For two capacitors in parallel, the charges divide as:

Q_{1} = \frac{C _{1}}{C _{1} + C _{2}} Q, Q_{2} = \frac{C _{2}}{C _{1} + C _{2}} Q

and the total energy stored is $U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$ .

Memory hook: Capacitors combine oppositely to resistors. Series capacitors add reciprocals (like parallel resistors); parallel capacitors add directly (like series resistors).

Solved Examples

Solved Example 1

In the circuit shown, four capacitors are arranged such that

C_{1} = 8 μ F

is in series with a combination where

C_{2} = 8 μ F

is in parallel with the series combination of

C_{3} = 8 μ F

and

C_{4} = 8 μ F

. A battery of

10 V

is applied. Find the potential difference across

A

and

B

(the two ends of the parallel branch of

C_{3}

C_{4}

Solution:

Label the capacitors 1, 2, 3, 4 for clarity. Capacitors 3 and 4 are in series:

C_{34} = \frac{C _{3} \cdot C _{4}}{C _{3} + C _{4}} = \frac{8 \times 8}{8 + 8} = 4 μ F

Then $C_{34}$ is in parallel with $C_{2}$ :

C_{2, 34} = 8 + 4 = 12 μ F

Finally $C_{1}$ is in series with this:

C_{e q} = \frac{8 \times 12}{8 + 12} = 4.8 μ F

Total charge drawn from the battery: $q = C_{e q} V = 4.8 \times 10 = 48 μ C$ .

Charge on $C_{1}$ is $48 μ C$ , so voltage across it: $V_{1} = q / C_{1} = 48/8 = 6 V$ .

Voltage across the parallel section $V_{P Q} = 10 - 6 = 4 V$ . By symmetry of $C_{3}$ and $C_{4}$ , the voltage across $A B$ (across just one of them) is $V_{A B} = 4/2 = 2 V$ .

Solved Example 2

In the bridge network shown, five identical capacitors each of value

C

are connected between terminals

A

and

B

: four form a diamond around the outside, and one bridges across the middle. Find the equivalent capacitance between

A

and

B

Figure: Symmetric bridge capacitor network. Middle capacitor may be dropped when its two ends are at equal potential.

Solution:

By symmetry of the network, the two junction points where the middle capacitor connects (let us call them $E$ and $D$ ) sit at the same potential. Since no charge flows through a capacitor whose two plates are at equal potential, the middle bridge capacitor can be effectively removed.

What remains: two parallel paths from $A$ to $B$ , each consisting of two capacitors in series. Each path has capacitance:

C_{path} = \frac{C \cdot C}{C + C} = \frac{C}{2}

The two paths are in parallel:

C_{A B} = \frac{C}{2} + \frac{C}{2} = C

Answer: $C_{e q} = C$ .

Solved Example 3

Two isolated conducting spheres of radii

R_{1} = 3 cm

and

R_{2} = 6 cm

carry charges

q_{1} = 6 μ C

and

q_{2} = 3 μ C

respectively. They are connected by a thin wire. Find (a) the common potential after connection and (b) the final charges on each sphere.

Solution:

Capacitance of each isolated sphere: $C_{1} = 4 π ε_{0} R_{1}$ and $C_{2} = 4 π ε_{0} R_{2}$ . So $C_{1} / C_{2} = R_{1} / R_{2} = 1/2$ .

(a) The common potential is:

V = \frac{q _{1} + q _{2}}{C _{1} + C _{2}} = \frac{9 μ C}{4 π ε _{0} ( R _{1} + R _{2} )} = \frac{9 \times 1 0 ^{- 6} \times 9 \times 1 0 ^{9}}{0.09} V = 9 \times 1 0^{5} V

(b) Final charges are in the ratio $C_{1} : C_{2} = R_{1} : R_{2} = 1 : 2$ . Total is $9 μ C$ , so:

q_{1}^{'} = 3 μ C, q_{2}^{'} = 6 μ C

Sphere 1 lost $3 μ C$ and sphere 2 gained $3 μ C$ .

Common Mistakes to Avoid

Watch out

Confusing series and parallel formulas: capacitors work oppositely to resistors. Series capacitors use the reciprocal-sum, parallel capacitors add directly.
Assuming charge on parallel-combined capacitors is equal. It is not - voltage is equal in parallel; charge is equal in series.
Forgetting the wire-connection charge sharing formula. Charges distribute in the ratio of capacitances ( $q_{1}^{'} / q_{2}^{'} = C_{1} / C_{2}$ ), not equally, not in the original ratio.
Missing the energy loss in charge sharing. When two capacitors at different potentials are connected, energy is always lost (unless $V_{1} = V_{2}$ ), even with a resistanceless wire.
Treating capacitance as depending on charge. Capacitance is a purely geometric and material property; increasing $Q$ increases $V$ proportionally, keeping $C$ fixed.

Frequently Asked Questions

Q1. What is the difference between a conductor's capacitance and a capacitor's capacitance?

A single isolated conductor has capacitance $C = Q / V$ where $V$ is measured from infinity. A capacitor has two conductors, and its capacitance $C = Q /Δ V$ uses the potential difference between the two plates. Capacitor values are much larger than isolated-conductor values for the same size, because the nearby second plate reduces the potential per unit charge.

Q2. Does capacitance depend on the material of the plates?

No. As long as the plates are conducting, capacitance depends only on the geometry (size, shape, plate separation) and on the dielectric medium between them, not on which specific metal is used. Copper and aluminum plates of the same geometry give identical capacitance.

Q3. Why is a farad such a large unit that we rarely see capacitors rated in whole farads?

One farad means storing $1 C$ of charge at just $1 V$ . To get $1 F$ from a parallel plate capacitor with a $1 mm$ gap, you would need plates about $110 km^{2}$ in area. Everyday electronics use $μ F$ , $n F$ or $pF$ ; supercapacitors that reach whole farads are physically large.

Q4. When two capacitors at different voltages are connected, where does the lost energy actually go?

Most of it becomes heat in the resistance of the connecting wire. If the wire is ideal (zero resistance), the transient current still radiates electromagnetic energy, and inductive effects cause oscillation that damps out. The amount lost, $\frac{1}{2} \cdot C_{1} C_{2} / (C_{1} + C_{2}) \cdot (V_{1} - V_{2})^{2}$ , is independent of the wire resistance - only the mechanism of loss changes.

Q5. In a series combination of capacitors, is the middle capacitor's charge really the same as the end ones?

Yes. In a series chain, the same current flows through the entire branch during charging. Since capacitor charge is the time-integral of current, and the current is common, each capacitor accumulates the same charge. This is why the smallest capacitor in a series combination sees the largest voltage drop.

Q6. How is a bridge capacitor network solved?

First check for symmetry: if the potentials at the two ends of the middle (bridge) capacitor are equal by circuit symmetry, that capacitor can be removed. The network then reduces to series-parallel combinations. If the bridge is unbalanced, use Kirchhoff's rules (charge conservation at junctions and voltage-loop equations) to solve.

Q7. Do JEE and NEET both ask capacitor combination problems?

Yes. JEE typically demands more calculation-heavy bridge networks and symmetry-exploitation problems, while NEET usually asks direct series-parallel simplification and formula-application questions. Both consistently include questions on charge sharing between two capacitors and energy loss.

Q8. Can capacitance be negative?

No. For a passive two-conductor system, adding charge always raises the potential difference in the same sense, so $C = Q / V$ is always positive. Effective negative capacitance appears only in active circuits (with amplifiers) and in some exotic ferroelectric materials - not in the standard JEE and NEET syllabus.

Previous year questions on Capacitors And Capacitance

19 questions from past papers, each with a step-by-step solution.

Show all 19 questions

Ready to master Capacitors?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test