Electric Field and Potential Relation Higher Difficulty Quiz

⚡ Electric Field & Potential – Advanced MCQs

1. \(V(x)=V_0(1 - \dfrac{x^2}{a^2})\). Where is electric field maximum?

x = 0 \(x = \dfrac{a}{2}\) x = a same everywhere

2. \(V = kr^3\). Electric field varies as:

r² r³ r constant

3. Equipotential: \(2x+3y+6z=const\). Direction of E?

2i+3j+6k -2i-3j-6k perpendicular cannot determine

4. \(E=(2x i +3j)\). Potential difference from (0,0) to (1,1)?

2.5 -4 -3 3

5. \(E=E_0 x\). Potential difference from 0 to a?

E₀a E₀a²/2 -E₀a²/2 -E₀a

6. Potential constant except at one point. Field is:

zero everywhere non-zero there undefined there cannot exist

7. Motion along equipotential surface. Work done?

max zero negative positive

8. \(V = k/r + c/r^2\). Field is:

k/r² + 2c/r³ k/r² - 2c/r³ -k/r² none

9. Which is NOT possible?

∮E·dl ≠ 0 E = -∇V conservative field path independent

10. If \(V ∝ 1/r^n\) and \(E ∝ 1/r^4\), find n.

2 3 4 5

Detailed Solutions

Q1: Given \( V(x)=V_0\left(1-\frac{x^2}{a^2}\right) \) Electric field: \[ E = -\frac{dV}{dx} = -V_0\left(-\frac{2x}{a^2}\right) = \frac{2V_0 x}{a^2} \] Thus \( E \propto x \), so it is maximum at maximum \( x \), i.e. \( x=a \). Answer: C

Q2: Given \( V = kr^3 \) \[ E = -\frac{dV}{dr} = -3kr^2 \] Magnitude: \[ |E| \propto r^2 \] Answer: A

Q3: Equipotential surface: \( 2x + 3y + 6z = \text{constant} \) \[ \nabla V = 2\hat{i} + 3\hat{j} + 6\hat{k} \] Electric field: \[ \vec{E} = -\nabla V = -(2\hat{i} + 3\hat{j} + 6\hat{k}) \] Answer: B

Q4: \[ V_B - V_A = -\int \vec{E} \cdot d\vec{r} \] \[ = -\int (2x\,dx + 3\,dy) \] \[ = -\left[\int_0^1 2x\,dx + \int_0^1 3\,dy\right] \] \[ = -(1 + 3) = -4 \] Answer: B

Q5: \[ V = -\int E\,dx = -\int_0^a E_0 x\,dx \] \[ = -E_0 \frac{a^2}{2} \] Answer: C

Q6: Electric field is: \[ E = -\frac{dV}{dx} \] At a point where potential is discontinuous, derivative is undefined. Answer: C

Q7: Work done: \[ W = -q\Delta V \] Along equipotential: \[ \Delta V = 0 \Rightarrow W = 0 \] Answer: B

Q8: \[ V = \frac{k}{r} + \frac{c}{r^2} \] \[ E = -\frac{dV}{dr} = -\left(-\frac{k}{r^2} - \frac{2c}{r^3}\right) \] \[ = \frac{k}{r^2} + \frac{2c}{r^3} \] Answer: A

Q9: Electrostatic field is conservative: \[ \oint \vec{E}\cdot d\vec{l} = 0 \] So non-zero loop integral is impossible. Answer: A

Q10: \[ V \propto \frac{1}{r^n} \] \[ E = -\frac{dV}{dr} \propto \frac{n}{r^{n+1}} \] Given: \[ E \propto \frac{1}{r^4} \Rightarrow n+1 = 4 \Rightarrow n = 3 \] Answer: B