Question

Two point charges of $+5\times {10}^{-19}C$ and $+20\times {10}^{-19}C$ are separated by a distance of $2m$. Find the point on the line joining them at which electric field intensity is zero.

Answer 1

Let charges $q_1=5\times {10}^{-19}C$ and $q_2=+20\times {10}^{-19}C$ be placed at $A$ and $B$ respectively. Distance $AB=2m$.
As charges are similar, the electric field strength will be zero between the charges on the line joining them. Let $P$ be the point (at a distance $x$ from $q_1)$ at which electric field intensity is zero. Then, $AP=xmetre,BP=(2-x)metre$. The electric field strength at $P$ due to charge $q_1$ is
${\overrightarrow{E}}_1=\frac{1}{4\pi {ϵ}_0}\frac{q_1}{x^2}$, along the direction $A$ to $P$.
The electric field strength at $P$ due to charge $q_2$ is
${\overrightarrow{E}}_2=\frac{1}{4\pi {ϵ}_0}\frac{q_2}{(2-x{)}^2}$, along the direction $B$ to $P$.
Clearly, ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ and are opposite in direction and for net electric field at $P$ to be zero, ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ and must be equal in magnitude.
So, $E_1=E_2$
$\Rightarrow \frac{1}{4\pi {ϵ}_0}\frac{q_1}{x^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_2}{(2-x{)}^2}$
Given, $q_1=5\times {10}^{-19}C,q_2=20\times {10}^{-19}C$
Therefore, $\frac{5\times {10}^{-19}}{x^2}=\frac{20\times {10}^{-19}}{(2-x{)}^2}$
or $\frac{1}{2}=\frac{x}{2-x}$
or $x=\frac{2}{3}m$.