Q4. Assertion: it is easier to find the potential of a set of charges than their field. Reason: potentials add as signed numbers, while fields add as vectors. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion (NCERT Eq. (2.17), page 51).
Q5. Charges of +2 µC, −3 µC and +4 µC sit at the corners of an equilateral triangle of side 0.30 m. Find the potential at a point 0.30 m from each (off the plane of the triangle), and at the midpoint of the side joining the +2 µC and −3 µC charges. Take k = 9 × 109 N m2 C−2.
At the equal-distance point: V = k(2 − 3 + 4) × 10−6/0.30 = 9.0 × 104 V. At the midpoint, the first two charges are 0.15 m away and the third is the triangle’s height, 0.30 × √3/2 ≈ 0.260 m, away. The three potentials are 1.20 × 105 V, −1.80 × 105 V and 1.39 × 105 V, so V ≈ 7.9 × 104 V (NCERT Eq. (2.18), page 52).
Q6. A uniformly charged thin spherical shell has a potential of 300 V at its surface, radius 5 cm. What is the potential at its centre, and at 15 cm from the centre?
300 V at the centre, the same as at the surface; at 15 cm, three times the radius, 300/3 = 100 V (NCERT Eq. (2.19), page 52).
Q7. A charge Q is spread unevenly round a thin ring of radius R. What is the potential at the centre of the ring?
Every bit of the charge is R from the centre, so the potentials of the bits add to V = Q/(4πε0R), however the charge is spread (reasoned from NCERT Eq. (2.18) and the continuous-charge rule, page 52).