Q4. Assertion: the field of a large uniformly charged sheet does not weaken as you move away from it. Reason: by Gauss’s law with a box through the sheet, E = σ/2ε, which contains no distance. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion, for points not too far away compared with the size of the sheet (NCERT page 35).
Q5. A uniformly charged thin spherical shell of radius 0.10 m carries 2.0 × 10−9 C. Find the field (a) at 0.05 m from its centre, (b) at 0.20 m from its centre. Take 1/4πε0 = 9 × 109 N m2 C−2.
(a) Zero: the point is inside the shell. (b) E = kq/r2 = 9 × 109 × 2.0 × 10−9/0.04 = 450 N/C, radially outward (NCERT Eqs. (1.34) and (1.35), pages 35–36).
Q6. A long wire carries 3.0 × 10−8 C/m. Find the field 0.30 m from it. Take 1/2πε0 = 1.8 × 1010 N m2 C−2.
E = λ/(2πε0r) = 1.8 × 1010 × 3.0 × 10−8/0.30 = 1.8 × 103 N/C, radially outward (NCERT Eq. (1.32), page 34).
Q7. In NCERT’s early model of the atom, where is the field zero?
At r = R and everywhere outside the atom, because the enclosed charge there is zero (NCERT Example 1.12, page 37).
Q8. A solid ball of radius R carries charge q spread uniformly through its volume. Find the field inside and outside it.
Outside (r > R), a Gaussian sphere encloses all of q, so E = q/(4πε0r2), as for a shell (NCERT summary, page 40). Inside (r < R), it encloses qr3/R3, so E × 4πr2 = qr3/(ε0R3), which gives E = qr/(4πε0R3). The field rises linearly from zero at the centre to q/(4πε0R2) at the surface, then falls as 1/r2. (This is the method of NCERT Example 1.12, pages 36–37.)