V due to a system of multiple Charges, Work done to assemble multiple charges to a given configuration, Work done or energy of a charge inside an external electric field, Work done or energy of two charges inside an external electric field. Quiz1

 

⚡ Worksheet: V due to a system of multiple Charges, Work done to assemble multiple charges to a given configuration, Work done or energy of a charge inside an external electric field, Work done or energy of two charges inside an external electric field. Quiz1

Section A: Theory (Conceptual Questions)

(1 mark each)

1. Define electric potential due to a system of point charges.
2. Is electric potential a scalar or vector? Explain briefly.
3. Write the principle used to find potential due to multiple charges.
4. What is meant by superposition of electric potential?
5. Can electric potential be zero while electric field is non-zero? Explain.
6. Define work done in assembling a system of charges.
7. What is electrostatic potential energy of a system?
8. When is work done positive while assembling charges?
9. Define external electric field.
10. What happens to potential energy when a charge moves against an electric field?

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Section B: Numerical Problems

(2–3 marks each)

Electric Potential due to Multiple Charges

11. Three charges $+2\mu C$, $+3\mu C$, and $-1\mu C$ are placed at the vertices of an equilateral triangle of side 1 m. Find the potential at the center.
12. Two charges $+5\mu C$ and $-5\mu C$ are separated by 2 m. Find the potential at the midpoint.
13. Four equal charges $+1\mu C$ are placed at the corners of a square of side 1 m. Find potential at the center.

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Work Done to Assemble Charges

14. Calculate the work required to assemble three charges $+2\mu C$, $+2\mu C$, and $+2\mu C$ at the corners of an equilateral triangle of side 1 m.
15. Two charges $+3\mu C$ and $+4\mu C$ are brought from infinity to a distance of 0.5 m. Find work done.
16. Four charges each $+1\mu C$ are placed at corners of a square of side 1 m. Find total electrostatic energy.

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Work Done in External Electric Field

17. A charge $2\mu C$ is moved through a potential difference of 100 V. Find work done.
18. A charge $5\mu C$ moves in a uniform electric field of $1000\ N/C$ over a distance of 2 m along the field. Find work done.

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Two Charges in External Field

19. Two charges $+2\mu C$ and $-2\mu C$ are placed in a uniform electric field of $500\ N/C$ separated by 0.1 m along field direction. Find total work done.
20. A dipole of charges $+q$ and $-q$ separated by distance $d$ is placed in a uniform electric field $E$. Find expression for work done in rotating it from $0^\circ$ to $90^\circ$.

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✅ Answer Key

1. Sum of potentials due to individual charges
2. Scalar
3. Superposition principle
4. Algebraic sum of potentials
5. Yes (example: dipole midpoint)
6. Work to bring charges from infinity
7. Stored energy due to configuration
8. Like charges
9. Field due to external sources
10. Increases
11. $V = 36 \times 10^3\ V$
    
12. $V = 0$
    
13. $V = 36 \times 10^3\ V$
    
14. $U = 0.108\ J$
    
15. $U = 0.216\ J$
    
16. $U = 0.072\ J$
    
17. $W = 2 \times 10^{-4}\ J$
    
18. $W = 0.01\ J$
    
19. $W = 0.2\ J$
    
20. $W = qEd$
    

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🧠 Detailed Solutions

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Q11 Solution

$$V = k \sum \frac{q}{r}$$

At center of equilateral triangle:

$$r = \frac{a}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$$$V = 9 \times 10^9 \times \frac{(2+3-1)\times 10^{-6}}{1/\sqrt{3}}$$ $$V = 9 \times 10^9 \times 4 \times 10^{-6} \times \sqrt{3}$$
$$V \approx 36 \times 10^3\ V$$

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Q12 Solution

Midpoint distances equal:

$$V = k\left(\frac{+q}{r} + \frac{-q}{r}\right) = 0$$

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Q13 Solution

Distance center to corner:

$$r = \frac{a}{\sqrt{2}}$$ $$V = 4 \times \frac{kq}{r} = 4 \times \frac{9 \times 10^9 \times 10^{-6}}{1/\sqrt{2}}$$ $$V = 36 \times 10^3\ V$$

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Q14 Solution

Energy of 3-charge system:

$$U = k\left(\frac{q^2}{a} + \frac{q^2}{a} + \frac{q^2}{a}\right) = 3 \frac{kq^2}{a}$$ $$U = 3 \times \frac{9 \times 10^9 \times (2\times10^{-6})^2}{1}$$ $$U = 0.108\ J$$

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Q15 Solution

$$U = \frac{kq_1 q_2}{r}$$ $$= \frac{9 \times 10^9 \times 3 \times 10^{-6} \times 4 \times 10^{-6}}{0.5}$$ $$U = 0.216\ J$$

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Q16 Solution

Total energy of square:

6 pairs exist:

$$U = 6 \times \frac{kq^2}{r}$$ $$U = 6 \times \frac{9 \times 10^9 \times 10^{-12}}{1} = 0.072\ J$$

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Q17 Solution

$$W = qV = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}\ J$$

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Q18 Solution

$$W = qEd$$
$$= 5 \times 10^{-6} \times 1000 \times 2 = 0.01\ J$$

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Q19 Solution

Net work:

$$W = qEd + (-q)E(d)$$
$$W = 2qEd = 2 \times 2 \times 10^{-6} \times 500 \times 0.1 = 0.2\ J$$

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Q20 Solution

Work done in rotation:

$$W = pE (\cos\theta_1 - \cos\theta_2)$$
$$= qdE (1 - 0) = qEd$$