Q4. Assertion: the electrostatic forces that two point charges exert on each other are equal and opposite. Reason: from Coulomb’s law in vector form, F = −F. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: interchanging 1 and 2 in NCERT’s Eq. (1.3) gives F = −F, so Coulomb’s law agrees with Newton’s third law (NCERT page 9).
Q5. Two point charges exert a force F on each other. Each charge is doubled and the distance between them is also doubled. What is the new force?
F′ = k(2q)(2q)/(2r) = 4kqq/4r = F: the force is unchanged, as in NCERT Example 1.4 (pages 10–11).
Q6. Two identical small metal spheres carry +4 µC and −2 µC and are 30 cm apart. They are touched together and put back at the same places. Find the force before and after.
Before: |q1q2| = 8 × 10−12 C2, so k|q1q2| = 7.2 × 10−2 N m2 and F = 7.2 × 10−2/0.09 = 0.8 N, attractive. Touching shares the total, +2 µC, equally, so each carries +1 µC. After: kq2 = 9 × 10−3 N m2 and F = 9 × 10−3/0.09 = 0.1 N, repulsive (NCERT pages 4–5 and 7–8).
Q7. What is the force between two small charged spheres with charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air? (NCERT Exercise 1.1, page 42)
q1q2 = 6 × 10−14 C2, so kq1q2 = 5.4 × 10−4 N m2; with r2 = 0.09 m2, F = 5.4 × 10−4/0.09 = 6 × 10−3 N, repulsive, since both charges are positive (air treated as vacuum).
Q8. The force on a small sphere of charge 0.4 µC due to another of charge −0.8 µC in air is 0.2 N. (a) What is the distance between them? (b) What is the force on the second sphere due to the first? (NCERT Exercise 1.2, page 42)
(a) k|q1q2| = 9 × 109 × 3.2 × 10−13 = 2.88 × 10−3 N m2, so r2 = k|q1q2|/F = 2.88 × 10−3/0.2 = 1.44 × 10−2 m2, and r = 0.12 m = 12 cm. (b) 0.2 N, attractive, pulling the second sphere towards the first: equal and opposite, by Newton’s third law (NCERT page 9).
Q9. Check that the ratio ke2/Gmemp is dimensionless, find its value, and say what it signifies (NCERT Exercise 1.3, page 42).
Units: (N m2 C−2)(C2)/[(N m2 kg−2)(kg)(kg)] = 1, so it is dimensionless. With k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, G = 6.67 × 10−11 N m2 kg−2, me = 9.11 × 10−31 kg and mp = 1.67 × 10−27 kg, it is about 2.3 × 1039 (NCERT prints 2.4 × 1039; these values give 2.3 × 1039). It is the ratio of the electric to the gravitational force between an electron and a proton, the same at every distance: the electric force is enormously stronger (NCERT Example 1.3, pages 9–10).
Q10. (a) Two insulated charged copper spheres A and B, with centres 50 cm apart, each carry 6.5 × 10−7 C. Their radii are negligible compared with the separation. What is the force of repulsion? (b) What is it if each charge is doubled and the distance halved? (NCERT Exercise 1.12, page 42)
(a) q2 = 4.23 × 10−13 C2 and r2 = 0.25 m2, so F = 9 × 109 × 4.23 × 10−13/0.25 ≈ 1.5 × 10−2 N. (b) Doubling both charges multiplies F by 4 and halving the distance multiplies it by 4 again: 16 × 1.52 × 10−2 ≈ 0.24 N.