Q: If two like charges of magnitude 1 × 10^{-9} coulomb and 9 × 10^{-9} coulomb are separated by a distance of 1 metre, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is:

(a) 0.25 m from the charge 1 × 10^{-9} coulomb

(b) 0.75 m from the charge 9 × 10^{-9} coulomb

(c)Both 1 and 2

(d) At all points on the line joining the charges

Ans: (c)

Sol: Let AB be the line joining the charges A (1 × 10^{-9} C) and B (9 × 10^{-9} C) . Let a charge q be placed at Point C & force acting on it is zero and AC = x , therefore BC = (1-x)

$F_{CA} = F_{CB}$

$\large \frac{1}{4 \pi \epsilon_0} \frac{q \times 1 \times 10^{-9}}{x^2}= \frac{1}{4 \pi \epsilon_0} \frac{q \times 9 \times 10^{-9}}{(1-x)^2} $

$\large \frac{1}{x^2} = \frac{9}{(1-x)^2} $

$\large \frac{1}{x} = \frac{3}{1-x} $

3x = 1-x ⇒ 4x = 1

x = 0.25 m