Question

i. ${\sigma }_a$	m. $\frac{q_a+q_b}{4\pi R^2}$
ii. ${\sigma }_b$	n. $\frac{-q_a}{4\pi a^2}$
iii. ${\sigma }_R$	o. $\frac{-q_b}{4\pi b^2}$

Match the table.

• A. (i, o), (ii, n), (iii, m)
• B. (i, n), (ii, o), (iii, m)
• C. (i, m), (ii, o), (iii, n)
• D. (i, n), (ii, m), (iii, o)

Answer 1

Answer: (B)

(i, n), (ii, o), (iii, m)

${\sigma }_A=\frac{{-q}_A}{4\pi a^2}{\sigma }_B=\frac{{-q}_b}{4\pi b^2}{\sigma }_R=\frac{q_A{+q}_B}{4\pi R^2}$

where,$a,b,R$ are

radii of surface $A,B\&R$ respectively

if surface $A\&B$ have $-ve$ charge, then surface are would have a positive charge.