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Class 12 Physics Chapter 1: Electric Charges and Fields MCQs with answers

10 multiple-choice questions on Electric Charges and Fields, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
Four small bodies inside an insulated box carry charges +4 μC, −7 μC, +2 μC and −1 μC. The total charge of the system is
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Answer: (d) −2 μC

Charge is a scalar and adds algebraically: (+4) + (−7) + (+2) + (−1) = −2 μC. Adding magnitudes gives 14 μC, which ignores the signs.

Question 2
Which of the following cannot be the charge on an isolated body? Take e = 1.6 × 10⁻¹⁹ C.
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Answer: (a) 2.4 × 10⁻¹⁹ C

Charge is quantised, q = ne with n an integer. 3.2, 4.8 and 6.4 (× 10⁻¹⁹ C) are 2e, 3e and 4e; 2.4 × 10⁻¹⁹ C is 1.5e, not a whole number of e.

Question 3
A glass rod is rubbed with a silk cloth and the rod becomes positively charged. During the rubbing
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Answer: (a) electrons move from the rod to the silk

In solids only some loosely bound electrons are transferred. The rod loses electrons to the silk, so the rod is left positive and the silk negative; no charge is created or destroyed.

Question 4
Take e = 1.6 × 10⁻¹⁹ C. 5.0 × 10¹² electrons are removed from a neutral metal sphere. The charge on the sphere is
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Answer: (a) +8.0 × 10⁻⁷ C

q = ne = 5.0 × 10¹² × 1.6 × 10⁻¹⁹ C = 8.0 × 10⁻⁷ C. Losing electrons leaves the sphere positive.

Question 5
Two point charges in vacuum exert a force F on each other. One charge is doubled and the distance between them is also doubled. The force becomes
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Answer: (b) F/2

F ∝ q₁q₂/r². Doubling one charge multiplies F by 2; doubling r divides it by 4. New force = 2F/4 = F/2.

Question 6
Point charges +2 μC and +8 μC are fixed 12 cm apart in vacuum. A third charge placed on the line between them feels no net force. Its distance from the +2 μC charge is
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Answer: (d) 4 cm

The two forces cancel when 2/x² = 8/(12 − x)², i.e. (12 − x)/x = 2, so x = 4 cm, nearer the smaller charge. 2.4 cm comes from dividing in the ratio of the charges instead of the square root of that ratio.

Question 7
Two equal point charges 10 cm apart in vacuum repel each other with a force of 0.9 N. (Take 1/4πε₀ = 9 × 10⁹ N m² C⁻².) The magnitude of each charge is
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Answer: (c) 1.0 μC

q² = Fr²/k = 0.9 × (0.10)² / (9 × 10⁹) = 1.0 × 10⁻¹² C², so q = 1.0 × 10⁻⁶ C = 1.0 μC.

Question 8
Two identical small conducting spheres carry +12 μC and −4 μC and attract each other with a force F when their centres are r apart. They are made to touch and are then put back r apart. The force between them now is
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Answer: (a) F/3, repulsive

Total charge is conserved: +12 − 4 = +8 μC, shared equally by identical spheres, so each has +4 μC. F ∝ |q₁q₂|: before 12 × 4 = 48, after 4 × 4 = 16. New force = 16F/48 = F/3, and like charges repel.

Question 9
For two protons, the ratio of the magnitude of the electric force to that of the gravitational force between them is R when they are 1 Å apart. When they are 2 Å apart the ratio is
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Answer: (b) R

Both forces vary as 1/r², so their ratio e²/(4πε₀Gmp²) does not depend on the separation. It stays R.

Question 10
Equal point charges +Q are fixed at two vertices of an equilateral triangle of side a in vacuum. The magnitude of the electric field at the third vertex is (k = 1/4πε₀)
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Answer: (b) √3 kQ/a²

Each charge gives kQ/a² along its own side, the two fields 60° apart. Resultant = 2(kQ/a²)cos 30° = √3 kQ/a², along the bisector, away from the two charges.

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