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

Electric Field - Potential Graph Based Concept Quiz

 

📊 Graph-Based MCQs (E–V Concepts)

---

Q1. (V vs x graph)

A graph of electric potential $V(x)$ vs $x$ is a straight line with negative slope.

Which statement is correct?

A) Electric field is zero
B) Electric field is constant and positive
C) Electric field is constant and negative
D) Electric field varies linearly

---

Q2. (V vs x parabola)

The graph of $V(x)$ vs $x$ is a parabola opening upward.

At the vertex of the parabola, electric field is:

A) maximum
B) zero
C) infinite
D) constant

---

Q3. (E vs x graph)

Electric field varies with position as shown:
A straight line passing through origin with positive slope in an $E$ vs $x$ graph.

Potential variation with $x$ is:

A) linear increasing
B) linear decreasing
C) quadratic decreasing (inverted parabola)
D) quadratic increasing (straight parabola)

---

Q4. (V vs r graph)

Potential $V(r)$ vs distance $r$ graph is a horizontal line.

Which is correct?

A) No electric field exists
B) Uniform electric field exists
C) Field is maximum
D) Field varies inversely with r

---

Q5. (V vs x piecewise graph)

Potential varies with position as follows:

• From $x=\mathrm{0\; to }$$x=a$: linear increase
• From $x=a$ to $x=2a$: constant
• From $x=2a$ to $x=3a$: linear decrease

Where is electric field zero?

A) only from $0 \to a$
B) only from $a \to 2a$
C) only from $2a \to 3a$
D) everywhere

---

✅ Answer Key

1. B
2. B
3. C
4. A
5. B

---

🧠 Detailed Solutions

---

Q1 Solution

$$E = -\frac{dV}{dx}$$

Slope is negative, so:

$$E = -(\text{negative}) = \text{positive constant}$$

✔ Answer: B

---

Q2 Solution

At vertex of parabola:

$$\frac{dV}{dx} = 0 \Rightarrow E = 0$$

✔ Answer: B

---

Q3 Solution

Given:

$$E \propto x$$
$$V = -\int E\,dx = -\int x\,dx = -\frac{x^2}{2}$$

So potential is quadratic decreasing (inverted parabola)

✔ Answer: C

---

Q4 Solution

$$E = -\frac{dV}{dr} = 0$$

✔ Answer: A

---

Q5 Solution

Electric field = slope of V– x graph

• Linear → constant field
• Horizontal → zero field

So only region $a \to 2a$

✔ Answer: B

---