Question

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

• A. $0.25m$ from the charge $1\times {10}^{-9}$ coulomb
• B. $0.75m$ from the charge $1\times {10}^{-9}$ coulomb
• C. Both $1$ and $2$
• D. at all points on the line joining the charges

Answer 1

The correct option is A $0.25m$ from the charge $1\times {10}^{-9}$ coulomb

Two like charges are of magnitude $Q_1=1\times {10}^{-9}$ coulomb and $Q_2=9\times {10}^{-9}$ coulomb respectively.

Let the point at which the force is zero be at $x$ metre distance from charge $Q_1$

So, distance from $Q_2$ is $(1-x)$ metre.

Force due to $Q_1$ = Force due to $Q_2$

$\begin{array}{c}\frac{K\times 1\times {10}^{-9}}{x^2}=\frac{K\times 9\times {10}^{-9}}{{\left(1-x\right)}^2}\\ {\left(1-x\right)}^2=9x^2\\ x=\frac{1}{4}m.\end{array}$

i.e. $0.25$m from the charge $1\times {10}^{-9}$C.