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Capacitance & Potential · NEET PYQs · Graph-based

Graph-based PYQs — Electrostatic Potential & Capacitance

Every graph-based NEET / AIPMT question from the chapter — V vs r, E vs r, Q vs V, U vs V, U vs Q, charging & discharging curves, C vs K, C vs d, force vs separation, equipotentials. For each: Question · Graph / circuit diagram · Given · Asked · Concept · Method · Easy trick · Solution.

1. V vs r — point charge 2. E vs r — point charge 3. V vs r — solid sphere 4. V vs r — hollow shell 5. E vs r — hollow shell 6. V vs r — concentric shells 7. Q vs V — slope is C 8. U vs V — parabola 9. U vs Q — parabola 10. C vs K 11. C vs d, C vs A 12. Q vs t — charging RC 13. I vs t — discharging 14. Force vs separation 15. Equipotentials
1

V vs r for a point charge

NEET PYQ
Q. Which graph correctly represents the variation of electric potential V with distance r from an isolated positive point charge?
(a) V ∝ r(b) V ∝ 1/r (hyperbola)(c) V ∝ 1/r²(d) constant
rVV = kq/r
Given
Isolated positive point charge q.
Asked
Shape of V(r).
Concept
For a point charge \(V = kq/r\). As r → 0, V → ∞; as r → ∞, V → 0. Curve is a rectangular hyperbola in the first quadrant.
Method
\[V(r) = \dfrac{kq}{r}\]
Plot: starts very high near origin, drops steeply, then flattens.
Easy trick
Point-charge V is a 1/r hyperbola; field E is a 1/r² steeper hyperbola. Don't confuse the two. For negative charge: same shape, mirrored below the r-axis.
Solution
V ∝ 1/r — hyperbolic curve. Answer: (b)
2

E vs r for a point charge

NEET PYQ
Q. Variation of electric field magnitude E with distance r from a point charge q:
(a) E ∝ r(b) E ∝ 1/r(c) E ∝ 1/r²(d) constant
rEE = kq/r²
Given
Point charge q.
Asked
Shape of E(r).
Concept
\(E = kq/r^2\). Curve falls off faster than V — at the same scale, E plummets to nearly zero quickly.
Method
\[E(r) = \dfrac{kq}{r^2}\]
Easy trick
Doubling r ⇒ V halves but E becomes one-quarter. "E falls off twice as fast as V" in log-log slope: V has slope −1, E has slope −2.
Solution
E ∝ 1/r². Answer: (c)
3

V vs r for a uniformly charged solid (insulating) sphere

NEET PYQ
Q. For a uniformly charged non-conducting solid sphere of radius R, total charge Q, the V vs r curve looks like:
(a) Constant inside, 1/r outside(b) Parabolic decrease inside (max at centre), 1/r outside(c) Zero inside, 1/r outside(d) Linear inside, linear outside
rVR3kQ/(2R)kQ/R
Given
Solid insulating sphere; charge Q distributed uniformly through volume.
Asked
V(r) shape both inside and outside.
Concept
Outside: same as a point charge → \(V = kQ/r\). Inside: \(V(r) = \dfrac{kQ}{2R^3}(3R^2 - r^2)\) — parabolic, maximum at centre = \(3kQ/(2R)\), drops to \(kQ/R\) at surface.
Method
\[V_{\text{centre}} = \dfrac{3 kQ}{2R},\quad V_{\text{surface}} = \dfrac{kQ}{R},\quad V_{\text{outside}} = \dfrac{kQ}{r}\]
Continuous at r = R (no jump). Curve is parabolic inside, hyperbolic outside.
Easy trick
Centre V of solid sphere = 1.5 × surface V. Memorise: "3/2 at the centre." For a hollow shell, inside V is constant = surface V (no 1.5 factor).
Solution
Parabolic decrease inside, 1/r outside. Answer: (b)
4

V vs r for a hollow conducting shell

NEET PYQ
Q. A hollow conducting spherical shell of radius R has charge Q. The V vs r graph is:
(a) Parabolic inside, 1/r outside(b) Constant (kQ/R) inside, 1/r outside(c) Zero everywhere inside(d) 1/r² inside, 1/r outside
rVRV = kQ/R (const)
Given
Hollow conducting shell with total charge Q on its surface.
Asked
V(r) shape.
Concept
Inside a charged shell E = 0 (Gauss). Therefore V is CONSTANT inside, equal to its surface value \(kQ/R\). Outside: as a point charge \(V = kQ/r\).
Method
\[V_{\text{inside}} = V_{\text{surface}} = \dfrac{kQ}{R},\quad V_{\text{outside}} = \dfrac{kQ}{r}\]
Graph: flat plateau till R, then 1/r decay.
Easy trick
"Hollow shell ⇒ V flat inside." Two graphs look alike outside (solid & hollow both 1/r), but inside the solid is a parabola peaking at 1.5× while the hollow is a flat plateau. That's the discriminator.
Solution
Constant kQ/R inside, kQ/r outside. Answer: (b)
5

E vs r for a hollow conducting shell

NEET 2018
Q. For a hollow conducting shell of radius R, charge Q, the magnitude of electric field E vs r:
(a) Zero inside, jumps to kQ/R² at surface, 1/r² outside(b) Constant inside(c) Linear inside, 1/r outside(d) 1/r² everywhere
rERkQ/R²E = 0 inside
Given
Hollow charged shell, radius R, charge Q.
Asked
E(r) shape.
Concept
Gauss: inside the shell (r < R) no enclosed charge ⇒ E = 0. Outside (r > R) it behaves like a point charge at centre.
Method
\[E(r) = \begin{cases} 0, & r < R \\ \dfrac{kQ}{r^2}, & r \ge R \end{cases}\]
A vertical jump at r = R from 0 to \(kQ/R^2\).
Easy trick
Hollow shell E graph: flat zero, then vertical jump at R, then 1/r² decay. V is continuous; E is discontinuous (because surface has charge density).
Solution
Zero inside, jump to kQ/R² at surface, 1/r² outside. Answer: (a)
6

V vs r for two concentric charged shells

NEET PYQ
Q. Inner shell radius a has charge +q; outer shell radius b has charge +Q. Sketch V(r). The potential of the inner shell is:
(a) k(q/a + Q/b)(b) kq/a + kQ/a(c) kq/b + kQ/a(d) kQ/(a+b)
rVabV_in
Given
Two concentric shells. Inner: radius a, charge q. Outer: radius b, charge Q.
Asked
V at inner shell + shape of curve.
Concept
Both shells contribute. Inside a shell, V due to that shell = its surface value. So at r = a: V = (kq/a) [self] + (kQ/b) [outer shell, evaluated at any inside point = kQ/b].
Method
  1. r < a: \(V = kq/a + kQ/b\) — flat.
  2. a < r < b: \(V = kq/r + kQ/b\) — drops as 1/r.
  3. r > b: \(V = k(q + Q)/r\) — drops as 1/r.
Easy trick
"Inside a shell, that shell behaves like a constant kQ/R contribution; outside, like a point charge." Just add the two contributions piecewise.
Solution
V at inner shell = \(kq/a + kQ/b\). Answer: corresponds to (a) with the matched-radius form.
7

Q vs V graph — slope is the capacitance

NEET PYQ
Q. Two capacitors A and B are plotted as Q vs V (charge vs voltage). A has steeper slope than B. Then:
(a) C_A > C_B(b) C_A < C_B(c) C_A = C_B(d) cannot decide
VQA (steep)B
Given
Q-V plot for two capacitors; A steeper than B.
Asked
Which has greater capacitance?
Concept
\(Q = CV\). Slope of Q-V graph = C.
Method
Steeper slope ⇒ larger C. So \(C_A > C_B\).
Easy trick
For Q-V graph: slope = C. For V-Q (swapped axes): slope = 1/C. Always check what's on which axis.
Solution
\(C_A > C_B\). Answer: (a)
8

U vs V graph — parabola through origin

NEET PYQ
Q. For a fixed capacitor, the stored energy U is plotted vs the applied voltage V. The graph is:
(a) Straight line through origin(b) Parabola opening upward through origin(c) Hyperbola(d) Horizontal line
VUU = ½CV²
Given
Capacitance C fixed; vary V.
Asked
Shape of U vs V.
Concept
\(U = \tfrac{1}{2} C V^2\). Parabola in V, passing through origin, slope zero at V = 0.
Method
\[U(V) = \tfrac{1}{2} C V^2\]
Steepness scales with C.
Easy trick
Doubling V quadruples U. If asked "energy ratio when V is tripled" → answer 9. The graph is a true parabola, not a hyperbola.
Solution
Upward parabola through origin. Answer: (b)
9

U vs Q graph — also a parabola

NEET PYQ
Q. For a fixed capacitor, U is plotted vs Q. The shape is:
(a) Straight line(b) Parabola through origin(c) Hyperbola(d) Step
QUU = Q²/(2C)
Given
Capacitance C fixed; vary Q.
Asked
U(Q) shape.
Concept
\(U = Q^2/(2C)\) — quadratic in Q, parabola through origin.
Method
\[U(Q) = \dfrac{Q^2}{2 C}\]
Steepness inversely related to C.
Easy trick
U is parabolic in both V and Q. Linear in only one quantity: try plotting \(\sqrt{U}\) vs V (or vs Q) — straight line.
Solution
Parabola through origin. Answer: (b)
10

C vs K (dielectric constant) graph

NEET style
Q. A parallel-plate capacitor has area A and gap d, with a dielectric of variable constant K. The graph of C vs K is:
(a) Hyperbola(b) Straight line through origin(c) Parabola(d) Constant horizontal line
KCC = K·(ε₀A/d)
Given
Parallel-plate capacitor with dielectric K.
Asked
Shape of C(K).
Concept
\(C = K \cdot \varepsilon_0 A/d\). Linear in K, passing through origin (K=0 is unphysical, but mathematical intercept is 0).
Method
\[C(K) = K \cdot C_0,\quad C_0 = \dfrac{\varepsilon_0 A}{d}\]
Slope = C₀.
Easy trick
C is directly proportional to K — never inverse. If a graph shows hyperbolic decay in K, it's wrong.
Solution
Straight line through origin. Answer: (b)
11

C vs d (hyperbola) and C vs A (straight line)

NEET concept
Q. For a parallel-plate capacitor with fixed area A and dielectric (no K), the graph of C vs d is:
(a) Straight line(b) Hyperbola (C ∝ 1/d)(c) Parabola(d) Constant
And for fixed d, the graph of C vs A is: straight line through origin.
dCC ∝ 1/dACC ∝ A
Given
Parallel plate cap. Vary d (A fixed) → first graph. Vary A (d fixed) → second graph.
Asked
Shapes.
Concept
\(C = \varepsilon_0 A/d\). Inverse in d, direct in A.
Method
\[C \propto \dfrac{1}{d}\quad\text{(hyperbola)};\quad C \propto A\quad\text{(line)}\]
Easy trick
"A in the numerator → linear with A; d in the denominator → hyperbola with d." K is also a numerator factor, so C is linear in K too.
Solution
C vs d: hyperbola (b). C vs A: straight line through origin.
12

Q vs t — charging through a resistor (RC circuit)

NEET PYQ
Q. A capacitor C is charged through a resistor R from a battery V₀. The Q(t) curve is:
(a) Linear ramp(b) \(Q_0\,(1 - e^{-t/RC})\) — saturates(c) Exponential decay(d) Step function
tQQ₀ = CV₀τ = RC0.63 Q₀
Given
Capacitor C in series with R and battery V₀; switch closed at t = 0.
Asked
Q(t) shape and value at t = τ.
Concept
Solution of Kirchhoff's voltage law gives exponential rise to saturation.
Method
\[Q(t) = C V_0 \left(1 - e^{-t/RC}\right)\]
At \(t = \tau = RC\): \(Q = 0.632\,Q_0\). At \(t = 5\tau\): essentially fully charged.
Easy trick
"63% in one time-constant, 95% in three, 99% in five." For NEET MCQ, recognize the saturating exponential shape and the kink at τ.
Solution
\(Q = Q_0(1 - e^{-t/RC})\). Answer: (b)
13

I vs t — discharging an RC circuit

NEET PYQ
Q. A charged capacitor is discharged through R. The current I(t) is:
(a) Constant(b) \(I_0\,e^{-t/RC}\) — exponential decay(c) \(I_0(1 - e^{-t/RC})\)(d) Linear decrease
tII₀τ = RC0.37 I₀
Given
Capacitor charged to \(V_0\), discharged through R at t = 0.
Asked
I(t).
Concept
I = V/R = (Q/C)/R, with Q decaying exponentially.
Method
\[I(t) = I_0\,e^{-t/RC},\quad I_0 = \dfrac{V_0}{R}\]
At t = τ: I = 0.37 I₀.
Easy trick
Discharging current and Q both decay as \(e^{-t/RC}\). At τ they fall to 37% (= 1/e). Same time-constant rules.
Solution
Exponential decay \(I_0 e^{-t/RC}\). Answer: (b)
14

Force between plates vs separation (Q fixed)

NEET concept
Q. An isolated parallel-plate capacitor with charge Q has plates pulled apart. The force F between the plates vs separation d:
(a) Decreases as 1/d(b) Decreases as 1/d²(c) Independent of d(d) Increases with d
dFF = Q²/(2ε₀A) — flat
Given
Charge Q on plates of area A; plates pulled apart (battery off).
Asked
F(d).
Concept
Each plate sits in the field of the OTHER plate, \(\sigma/(2\varepsilon_0)\). Force = \(Q \cdot \sigma/(2\varepsilon_0) = Q^2/(2\varepsilon_0 A)\) — depends only on charge and area, NOT on d.
Method
\[F = \dfrac{Q^2}{2\varepsilon_0 A}\;\;(\text{independent of } d)\]
Easy trick
"Force between fixed-Q plates = same at any separation." But if the battery is on instead, F depends on d because Q is no longer fixed. Be alert which case.
Solution
F is independent of d. Answer: (c)
15

Equipotential surfaces — point charge vs parallel plates

NEET PYQ
Q. Identify the equipotential surfaces:
(i) Isolated point charge
(ii) Uniform field (parallel plates)
(iii) Electric dipole (at large r)
(i) Point charge(ii) Uniform fieldparallel planes(iii) Dipole
Given
Three standard configurations.
Asked
Shape of equipotential surfaces.
Concept
Equipotentials are surfaces of constant V, always ⊥ to E lines. They reflect the symmetry of the source.
Method
  1. Point charge: V = kq/r ⇒ spherical surfaces concentric with charge.
  2. Uniform field: E along x ⇒ V = −Ex ⇒ equipotentials are planes ⊥ to E.
  3. Dipole: complicated — distorted "egg-shaped" curves; midplane (perpendicular bisector) is the V = 0 surface.
Easy trick
"Equipotentials are perpendicular to field lines." So if field lines are radial → equipotentials are spheres. If field lines are straight parallel → equipotentials are parallel planes.
Solution
(i) Concentric spheres. (ii) Parallel planes ⊥ to field. (iii) Distorted closed curves; perpendicular-bisector plane = 0 V.