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

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Solution

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 $A B = 2 m$ .
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, $A P = x m e t r e, B P = (2 - x) m e t r e$ . The electric field strength at $P$ due to charge $q_{1}$ is
${\to E}_{1} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{x^{2}}$ , along the direction $A$ to $P$ .
The electric field strength at $P$ due to charge $q_{2}$ is
${\to E}_{2} = \frac{1}{4 π ϵ_{0}} \frac{q_{2}}{(2 - x)^{2}}$ , along the direction $B$ to $P$ .
Clearly, ${\to E}_{1}$ and ${\to E}_{2}$ and are opposite in direction and for net electric field at $P$ to be zero, ${\to E}_{1}$ and ${\to E}_{2}$ and must be equal in magnitude.
So, $E_{1} = E_{2}$
$\Rightarrow \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{x^{2}} = \frac{1}{4 π ϵ_{0}} \frac{q_{2}}{(2 - x)^{2}}$
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 $x = \frac{2}{3} m$ .

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Two point charges of +5×10−19C and +20×10−19C are separated by a distance of 2 m. Find the point on the line joining them at which electric field intensity is zero.

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