Equipotential Surfaces Equipotential surfaces due to point charge, in uniform external electric field, in a dipole, in a region of two charges of same polarity, Electric field perpendicular to equipotential surface, Equipotential surfaces in a non-uniform electric field. Quiz1

 

📘 Worksheet: Equipotential Surfaces Equipotential surfaces due to point charge, in uniform external electric field, in a dipole, in a region of two charges of same polarity, Electric field perpendicular to equipotential surface, Equipotential surfaces in a non-uniform electric field. Quiz1

🔹 Section A: Theory / Concept Questions (1–10)

1. Define an equipotential surface. Give one real-life example.
2. Why is no work done when a charge moves along an equipotential surface?
3. What is the shape of equipotential surfaces for a point charge? Explain.
4. Draw and describe equipotential surfaces in a uniform electric field.
5. How are equipotential surfaces oriented with respect to electric field lines?
6. Describe equipotential surfaces of an electric dipole.
7. What happens to the spacing between equipotential surfaces in regions of:
   • (a) Strong electric field
   • (b) Weak electric field
8. Explain equipotential surfaces for two like charges placed near each other.
9. Can two equipotential surfaces intersect? Justify your answer.
10. Describe equipotential surfaces in a non-uniform electric field with an example.

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🔹 Section B: Numerical Problems (11–20)

11. A charge $+5 \, \mu C$ creates a potential of $100 \, V$ at a point.
    Find the distance of that point from the charge.

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12. Two equipotential surfaces are at $50 \, V$ and $70 \, V$, separated by $0.2 \, m$.
    Find the magnitude of electric field between them.

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13. A charge moves along an equipotential surface.
    If the charge is $2 \, C$, find the work done.

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14. The electric field between two parallel plates is $500 \, N/C$.
    Find the potential difference between two equipotential surfaces $4 \, cm$ apart.

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15. Equipotential surfaces are closer together in a region.
    If the distance between surfaces is reduced to half, how does the electric field change?

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16. A point charge produces potential $V = 200 \, V$ at distance $r$.
    What will be the potential at distance $2r$?

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17. Two like charges create equipotential surfaces.
    At a point where potential is zero, explain if electric field can be non-zero.

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18. Potential changes from $10 \, V$ to $30 \, V$ over a distance of $0.1 \, m$.
    Find electric field.

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19. A dipole creates potential $V = 0$ at some point.
    Is electric field zero there? Explain numerically or conceptually.

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20. In a non-uniform field, potential changes from $100 \, V$ to $40 \, V$ over $0.3 \, m$.
    Find average electric field.

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✅ Answer Key

Section A (Theory)

1. Surface with constant potential
2. No potential difference → no work
3. Spherical surfaces
4. Parallel planes
5. Always perpendicular
6. Complex, symmetric around dipole axis
7. (a) Close → strong field (b) Far → weak field
8. Distorted shapes, no simple symmetry
9. No, otherwise two potentials at one point
10. Irregular surfaces (e.g., near multiple charges)

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Section B (Numerical)

11. $r = \dfrac{kq}{V} = \dfrac{9\times10^9 \times 5\times10^{-6}}{100} = 450 \, m$
    
12. $E = \dfrac{\Delta V}{d} = \dfrac{20}{0.2} = 100 \, N/C$
    
13. $W = 0$
    
14. $V = Ed = 500 \times 0.04 = 20 \, V$
    
15. Electric field doubles
16. $V \propto \dfrac{1}{r} \Rightarrow V = 100 \, V$
    
17. Yes, electric field can be non-zero
18. $E = \dfrac{20}{0.1} = 200 \, N/C$
    
19. No, electric field may be non-zero
20. $E = \dfrac{60}{0.3} = 200 \, N/C$
    

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🧠 Detailed Solutions

Q11

Using:

$$V = \frac{kq}{r} \Rightarrow r = \frac{kq}{V}$$

Substitute values:

$$r = \frac{9\times10^9 \times 5\times10^{-6}}{100} = 450 \, m$$

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Q12

$$E = \frac{\Delta V}{d} = \frac{70 - 50}{0.2} = 100 \, N/C$$

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Q13

Work done:

$$W = q \Delta V$$

Since $\Delta V = 0$,

$$W = 0$$

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Q14

$$V = Ed = 500 \times 0.04 = 20 \, V$$

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Q15

$$E \propto \frac{1}{d}$$

If distance halves → electric field doubles.

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Q16

$$V \propto \frac{1}{r}$$

At $2r$, potential becomes:

$$V = \frac{200}{2} = 100 \, V$$

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Q17

At zero potential point:

• Contributions cancel
• But field depends on vector sum → can be non-zero

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Q18

$$E = \frac{30 - 10}{0.1} = 200 \, N/C$$

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Q19

For dipole:

• Potential can be zero at equatorial line
• Electric field is not zero

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Q20

$$E = \frac{100 - 40}{0.3} = 200 \, N/C$$