Electrostatic Potential and Capacitance — Questions & Answers

Sample questions from Electrostatic Potential and Capacitance

1. Q1. The relationship E = -ΔV/Δr means:

   • A.Field points from low to high V
   • B.Field points from high to low V✓
   • C.Field is perpendicular to V
   • D.Field equals V
   Solution

   Negative sign: field points from HIGH potential to LOW potential (i.e., in direction of DECREASING potential).

2. Q2. For a parallel plate capacitor with plate area A and separation d (vacuum between plates), the capacitance is:

   • A.epsilon0 d / A
   • B.epsilon0 A / d✓
   • C.epsilon0 A d
   • D.A / (epsilon0 d)
   Solution

   C = epsilon0 A/d -- capacitance increases with larger plate area (more charge can be stored at the same field) and decreases with larger separation (weaker field, lower charge for the same voltage). A student who inverts the A and d dependence picks A.

3. Q3. The electric potential on the axis of a uniformly charged disc of radius R and surface charge density sigma, at a distance x from the centre, is V = (sigma/(2 eps0))(sqrt(x^2 + R^2) - x). The electric field on the axis at distance x is therefore:

   • A.(sigma/(2 eps0)) sqrt(x^2 + R^2)
   • B.(sigma/(2 eps0))(x/sqrt(x^2 + R^2) - 1)
   • C.(sigma/(2 eps0))(1 - x/sqrt(x^2 + R^2))✓
   • D.sigma/(2 eps0) independent of x
   Solution

   E = -dV/dx = -(sigma/(2 eps0))(x/sqrt(x^2 + R^2) - 1) = (sigma/(2 eps0))(1 - x/sqrt(x^2 + R^2)). At x = 0 this gives the infinite-sheet value sigma/(2 eps0), and for x >> R it falls off like a point charge.

4. Q4. Three identical capacitors, each of capacitance C, are connected in a triangle-like loop (capacitor 1 between nodes A-B, capacitor 2 between B-C, capacitor 3 between C-A), with no external connection made to any node. The equivalent capacitance of this isolated loop, as 'seen' between any pair of nodes if measured, is:

   • A.(3/2)C, the same value between any pair of nodes by symmetry✓
   • B.Exactly C, regardless of which two nodes are chosen
   • C.Always 3C between any pair
   • D.Always C/3 between any pair
   Solution

   By the full symmetry of the triangle (all three capacitors identical), between any chosen pair of nodes (say A and B), the network reduces to capacitor 1 directly between A-B (value C), in PARALLEL with the SERIES combination of capacitor 3 (C, between C-A) and capacitor 2 (C, between B-C), which forms an alternate A-to-B path through node C with series value C/2. Total: C_eq = C + C/2 = (3/2)C. By the triangle's full symmetry, this same (3/2)C value applies between ANY of the three node pairs. The key conceptual point tested here is that in an isolated (no external connection) symmetric network, the equivalent capacitance BETWEEN any given pair of terminals is still a well-defined, calculable quantity, even with no external circuit attached -- a student who assumes an unconnected loop has no meaningful 'equivalent capacitance' at all misses that this is simply asking what a meter WOULD read if connected across that pair.

5. Q5. The relationship between electric field E and potential V along a direction x is:

   • A.E = dV/dx
   • B.E = -dV/dx✓
   • C.E = -V/x always
   • D.V = -dE/dx
   Solution

   The field is the negative gradient of the potential, E = -dV/dx, meaning the field points in the direction of DEcreasing potential. A student who forgets the negative sign (thinking field points towards increasing potential) picks A.

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