Question

Two like charges of magnitude $1\times 10^{-9}$ coulomb and $9\times 10^{-9}$ coulomb are separated by a distance of $1meter.$ The point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is

• A. $0.25m$ from the charge $1\times 10^{-9}$ $C$
• B. $0.75m$ from the charge $9\times 10^{-9}$ $C$
• C. $0.25m$ from the charge $1\times 10^{-9}$ $C$ , $0.75m$ 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 $\frac{kq{q}_1}{x^2}=\frac{kq{q}_2}{{\left(1-x\right)}^2}$
or $\frac{1\times 10^{-9}}{x^2}=\frac{9\times 10^{-9}}{(1-x{)}^2}$
or $\frac{1}{x}=\frac{3}{1-x}$
or $1-x=3x$
or $x=0.25m$ from $1\times 10^{-9}C$
$\therefore$ From $9\times 10^{-9}C$ ,
distance $=1-x$
$=1-0.25$
$=0.75$