Question

The electric field intensity at $P$ and $Q$, in the shown arrangement, are in the ratio

• A. 1 : 2
• B. 2 : 1
• C. 1 : 1
• D. 4 : 3

Answer 1

Answer: (C)

$E_p=\frac{kq}{r^2}...(i)$
$E_Q=\frac{kq}{(2r{)}^2}+\frac{k\cdot 3q}{(2r{)}^2}$
$=\frac{kq}{4r^2}+\frac{k\cdot 3q}{4r^2}$
$=\frac{kq}{r^2}...(ii)$
$E_P:E_Q=1:1$