Question

Two like charges of magnitude $1\times 10^{- 9}$ coulomb and $9\times 10^{- 9}$ coulomb are separated by a distance of $1 \, meter.$ 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 $1\times 10^{- 9}$ $C$
• B. $0.75 \, m$ from the charge $9\times 10^{- 9}$ $C$
• C. $0.25 \, m$ from the charge $1\times 10^{- 9}$ $C$ , $0.75 \, m$ from the charge $9\times 10^{- 9}$ $C$
• D. at all points on the line joining the charges

Answer 1

Answer: (C)

$\because \, $ Force on charge $\text{q }$ is zero
$\therefore \, F_{21}=F_{23}$
or $\frac{k q q_{1}}{x^{2}}=\frac{k q q_{2}}{\left(1 - x\right)^{2}}$
or $\quad \frac{1 \times 10^{-9}}{x^2}=\frac{9 \times 10^{-9}}{(1-x)^2}$
or $\frac{1}{x}=\frac{3}{1 - x}$
or $1-x=3x$
or $x=0.25 \, m$ from $1\times 10^{- 9 \, }C$
$\therefore $ From $9\times 10^{- 9}C$ ,
distance $=1-x$
$=1-0.25$
$=0.75$