Question

A point charge $A$ of charge $+4 \mu$C and another point charge B of charge $-1 \mu\,C$ are placed in air at a distance $1\, metre$ apart. Then the distance of the point on the line joining the charges and from the charge $B$, where the resultant electric field is zero, is (in metre)

• A. 0.5
• B. 1.5
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Here $\frac{4 \times 10^{-6} }{(1 + x)^2} = \frac{10^{-6}}{x^2}$
$i.e.$ $\frac{4}{1} = \frac{(1 + x)^2}{x^2} \, i.e. \, \frac{2}{1} = \frac{1 + x}{x} \, i.e. \, 2x = 1 + x \, i.e.\, x = 1 \, m$