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EAPCET Physics · Intermediate second year

Electrostatic Potential and Capacitance

This chapter covers electric potential and potential energy, capacitors and their combinations, dielectrics and the energy stored in a capacitor.

Subject
Physics
Taught in
Intermediate second year
Physics in the paper
40 of 160 questions

Key formulas and results

  • Potential of a point charge: V=kqrV = \frac{kq}{r}
  • E=−dVdrE = -\frac{dV}{dr}
  • Parallel-plate capacitor: C=ε0AdC = \frac{\varepsilon_{0}A}{d}; with a dielectric: C=Kε0AdC = \frac{K\varepsilon_{0}A}{d}
  • Series: 1C=∑1Ci\frac{1}{C} = \sum \frac{1}{C_{i}}; parallel: C=∑CiC = \sum C_{i}
  • Energy U=12CV2=Q22CU = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}
  • Potential of a short dipole: V=kpcos⁡θr2V = \frac{kp\cos\theta}{r^{2}}; V=kpr2V = \frac{kp}{r^{2}} on the axis and V=0V = 0 on the equatorial line
  • Uniformly charged solid (non-conducting) sphere, inside (r<Rr < R): E=kQrR3E = \frac{kQr}{R^{3}} and V=kQ(3R2−r2)2R3V = \frac{kQ(3R^{2} - r^{2})}{2R^{3}}; at the centre V=3kQ2RV = \frac{3kQ}{2R}
  • Potential on the axis of a charged disc: V=σ2ε0[R2+x2−x]V = \frac{\sigma}{2\varepsilon_{0}}\left[\sqrt{R^{2} + x^{2}} - x\right]; at the centre V=σR2ε0V = \frac{\sigma R}{2\varepsilon_{0}}
  • Potential energy of a system of point charges: U=k∑qiqjrijU = k\sum \frac{q_{i}q_{j}}{r_{ij}}, summed over every pair once
  • Parallel-plate capacitor with a dielectric slab of thickness tt: C=ε0Ad−t+t/KC = \dfrac{\varepsilon_{0}A}{d - t + t/K}
  • Spherical capacitor: C=4πε0abb−aC = \frac{4\pi\varepsilon_{0}ab}{b - a}; isolated sphere: C=4πε0RC = 4\pi\varepsilon_{0}R; cylindrical capacitor: C=2πε0lln⁡(b/a)C = \frac{2\pi\varepsilon_{0}l}{\ln(b/a)}
  • Energy density of an electric field: u=12ε0E2u = \tfrac{1}{2}\varepsilon_{0}E^{2}; force between the plates of a capacitor: F=Q22Aε0F = \frac{Q^{2}}{2A\varepsilon_{0}}
  • Charging an RC circuit: q=q0(1−e−t/RC)q = q_{0}\left(1 - e^{-t/RC}\right), i=VRe−t/RCi = \frac{V}{R}e^{-t/RC}; after one time constant (t=RCt = RC) q=0.632q0q = 0.632q_{0}

What this chapter covers

  • Electric potential and potential energy
  • Equipotential surfaces
  • Capacitance
  • Combinations of capacitors
  • Dielectrics
  • Energy stored in a capacitor

How it is usually asked

  • Capacitor combinations
  • Energy stored
  • Effect of inserting a dielectric
  • Potential due to a system of charges

Understanding Electrostatic Potential and Capacitance

Electric potential is the work done per unit charge in bringing a positive test charge from infinity to a point. Unlike the field, it is a scalar, so potentials from several charges simply add with their signs. The field points in the direction of decreasing potential, E = −dV/dr, and no work is needed to move a charge along an equipotential surface.

A capacitor stores charge and energy. Its capacitance C = Q/V depends only on geometry and the medium; for a parallel-plate capacitor, C = ε₀A/d, multiplied by K when a dielectric fills the gap. Capacitors in parallel add directly, while in series their reciprocals add, the opposite of resistors. In series they carry the same charge; in parallel they share the same voltage.

EAPCET often tests capacitor combinations, stored energy, dielectric insertion and potential due to a set of charges. Keep the two dielectric cases apart. With the battery connected, V is fixed, so inserting a dielectric of constant K multiplies C, Q and the stored energy by K. With the battery removed, Q is fixed, so V falls to V/K and the stored energy also falls to 1/K of its value.

Worked example

EAPCET-style question

Capacitors of 2 μF, 3 μF and 6 μF are connected in series across a 12 V battery. The charge on each capacitor is

  1. (1)12 μC
  2. (2)132 μC
  3. (3)24 μC
  4. (4)6 μC

Solution

  1. Series combination: 1C=12+13+16=3+2+16=1\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1, so C=1 μFC = 1\ \mu\text{F}.
  2. Total charge Q=CV=1 μF×12 V=12 μCQ = CV = 1\ \mu\text{F} \times 12\ \text{V} = 12\ \mu\text{C}.
  3. In series every capacitor carries the same charge.
  4. Charge on each capacitor = 12 μC.

Answer: 12 μC

Shortcuts and results to remember

  • When nn identical charged drops (each at potential VV) merge into one drop, the new potential is n2/3Vn^{2/3}V: 27 drops give 9V9V and 64 drops give 16V16V.
  • Two conducting spheres joined by a wire reach the same potential, so the total charge divides in proportion to the radii: q1/q2=r1/r2q_{1}/q_{2} = r_{1}/r_{2}.
  • A conductor is an equipotential: inside a charged conducting sphere or shell E=0E = 0 and V=kQ/RV = kQ/R everywhere.
  • Potential at the centre of any arc or ring of total charge qq and radius RR is kq/Rkq/R whatever the arc’s shape; a semicircular ring of density λ\lambda gives V=πkλV = \pi k\lambda.
  • For concentric shells, the potential of a shell =k(total charge inside or on it)its radius+∑k(charge on each outer shell)that shell’s radius= \dfrac{k(\text{total charge inside or on it})}{\text{its radius}} + \sum \dfrac{k(\text{charge on each outer shell})}{\text{that shell’s radius}}; earthing a shell makes its potential zero.

More practice questions

Practice question 1

Two conducting spheres of radii 2r2r and 7r7r carry charges −40 μC and +76 μC. They are connected by a thin wire. The final charges on the smaller and larger spheres are

  1. (1)8 μC and 28 μC
  2. (2)18 μC and 18 μC
  3. (3)28 μC and 8 μC
  4. (4)−8 μC and 44 μC

Solution

  1. Total charge is conserved: −40+76=36 μC-40 + 76 = 36\ \mu\text{C}
  2. After connection both spheres are at the same potential, so q1/r1=q2/r2q_{1}/r_{1} = q_{2}/r_{2} and q1:q2=2:7q_{1} : q_{2} = 2 : 7
  3. q1=36×29=8 μCq_{1} = 36 \times \frac{2}{9} = 8\ \mu\text{C}
  4. q2=36×79=28 μCq_{2} = 36 \times \frac{7}{9} = 28\ \mu\text{C}

Answer: 8 μC and 28 μC

Practice question 2

A ring of radius RR carries charge QQ. The work done by an external agent to bring a charge qq slowly from a point on the axis at distance 3R\sqrt{3}R from the centre to the centre of the ring is

  1. (1)kQqR\frac{kQq}{R}
  2. (2)kQq2R\frac{kQq}{2R}
  3. (3)kQq3R\frac{kQq}{3R}
  4. (4)2kQqR\frac{2kQq}{R}

Solution

  1. Every point of the ring is at distance R2+3R2=2R\sqrt{R^{2} + 3R^{2}} = 2R from the axial point, so VA=kQ2RV_{A} = \frac{kQ}{2R}
  2. At the centre every point is at distance RR, so VC=kQRV_{C} = \frac{kQ}{R}
  3. Wext=q(VC−VA)=q(kQR−kQ2R)W_{\text{ext}} = q(V_{C} - V_{A}) = q\left(\frac{kQ}{R} - \frac{kQ}{2R}\right)
  4. Wext=kQq2RW_{\text{ext}} = \frac{kQq}{2R}

Answer: kQq2R\frac{kQq}{2R}

Common mistakes to avoid

  • Combining capacitors like resistors — for capacitors, parallel values add and series values add as reciprocals.
  • Adding potentials as vectors or ignoring their signs — potential is a scalar, and negative charges give negative potential.
  • Assuming zero field means zero potential (or the reverse) — at the midpoint of two equal like charges E = 0 but V ≠ 0.
  • Keeping V fixed after the battery is disconnected — with the battery removed, Q stays constant and V changes when a dielectric is inserted.

How to prepare it

When a dielectric is inserted, first ask whether the battery is still connected — V stays fixed if it is, Q stays fixed if it is not.

Once the formulas feel familiar, test them the way the exam does: practise Electrostatic Potential and Capacitance questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Electrostatic Potential and Capacitance chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Electrostatic Potential and Capacitance part of the EAPCET Physics syllabus?
Yes. Electrostatic Potential and Capacitance is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Electrostatic Potential and Capacitance for EAPCET?
When a dielectric is inserted, first ask whether the battery is still connected — V stays fixed if it is, Q stays fixed if it is not.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Electrostatic Potential and Capacitance?
Combining capacitors like resistors — for capacitors, parallel values add and series values add as reciprocals. Adding potentials as vectors or ignoring their signs — potential is a scalar, and negative charges give negative potential. Assuming zero field means zero potential (or the reverse) — at the midpoint of two equal like charges E = 0 but V ≠ 0. Keeping V fixed after the battery is disconnected — with the battery removed, Q stays constant and V changes when a dielectric is inserted.