Question

Two point charges $q_1=+9\mu C$ and $q_2=-1\mu C$ are held $10\text{cm}$ apart. Where should at third charge $+Q$ be placed from $q_2$ on the line joining them so that charge $Q$ does not experience any net force ?

• A. $4\text{cm}$
• B. $5\text{cm}$
• C. $6\text{cm}$
• D. $7\text{cm}$

Answer 1

The correct option is B $5\text{cm}$


It follows from figure that charge $Q$ will experience net force in the X-direction if it lies at any point between $AB$.

Let $x$ be the distance of $Q$ from $q_2$. Then forces exerted on $Q$ by $q_1$ and $q_2$ respectively are

$F_1=\frac{q_{1}Q}{4\pi {ϵ}_0(0.1+x{)}^2}=\frac{9\times {10}^{-6}Q}{4\pi {ϵ}_0(0.1+x{)}^2}$

and $F_2=\frac{q_{2}Q}{4\pi {ϵ}_0{x}^2}=\frac{-1\times {10}^{-6}Q}{4\pi {ϵ}_0{x}^2}$

Net force on $Q=F_1+F_2$
Net force on $Q$ is zero if $F_1+F_2=0$

$\Rightarrow \frac{9\times {10}^{-6}Q}{4\pi {ϵ}_0(0.1+x{)}^2}-\frac{1\times {10}^{-6}Q}{4\pi {ϵ}_0{x}^2}=0$

$\Rightarrow 9=\frac{(0.1+x{)}^2}{x^2}$

$\Rightarrow 3=\frac{0.1+x}{x}$ (ignore negative sign, as we get negative value of $x$)

$\Rightarrow x=0.05\text{m}=5\text{cm}$