Q1. What is the potential difference between two points in an electric field if the electric field strength is 50 N/C and the distance between the points is 0.4 m? (2023)
Solution:
Potential difference V = E * d = 50 N/C * 0.4 m = 20 V.
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Q2. If a charge of +10 μC is placed in an electric field of strength 200 N/C, what is the force experienced by the charge? (2019)
Solution:
Force F = qE = (10 × 10^-6 C) * (200 N/C) = 2 N.
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Q3. A charge of -2 µC is placed in an electric field of 500 N/C. What is the direction of the force acting on the charge? (2023)
Solution:
The force on a negative charge is in the direction opposite to the electric field, hence towards the positive charge.
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Q4. What happens to the electric field when the charge is doubled while keeping the distance constant? (2023)
Solution:
The electric field E is directly proportional to the charge, so if the charge is doubled, the electric field also doubles.
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Q5. If two charges of +2 µC and -2 µC are placed 1 m apart, what is the net electric field at the midpoint? (2023)
Solution:
The electric fields due to both charges at the midpoint cancel each other out, resulting in a net electric field of 0 N/C.
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Q6. What is the electric field due to a point charge of +10 µC at a distance of 0.1 m? (2023)
Solution:
Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (0.1 m)² = 9000 N/C.
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Q7. A charge of +10 µC is placed in an electric field of strength 2000 N/C. What is the force acting on the charge? (2023)
Solution:
Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.
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Q8. If the charge on a capacitor is doubled while keeping the voltage constant, what happens to the capacitance? (2022)
Solution:
Capacitance C = Q/V. If Q is doubled and V remains constant, C must also double.
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Q9. Which of the following is a property of electric field lines? (2023)
Solution:
Electric field lines cannot intersect because it would imply two different directions of the electric field at a point.
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Q10. What is the potential difference between two points in an electric field of 400 N/C separated by 0.5 m? (2022)
Solution:
Potential difference V = E * d = 400 N/C * 0.5 m = 200 V.
