The point on the line joining the two point charges be given by,
Given: Charge q_{1}=5 \times 10^{-19} C
Charge q_{2}=20 \times 10^{-19} C
Distance between the point charges q_{1} and q_{2} = 2 m
Let us assume the point at which electric field intensity is zero be at a distance from q_{2} = x
And from q_{1} = 2-x
As given that \bold{E_{1}-E_{2}=0}
\begin{array}{l}{E_{1}=E_{2}} \\ \\{E=\frac{K q}{r^{2}}} \\ \\{\frac{K q_{1}}{r_{1}^{2}}=\frac{K q_{2}}{r_{2}^{2}}} \\ \\{\frac{q_{1}}{r_{1}^{2}}=\frac{q_{2}}{r_{2}^{2}}}\end{array}
\begin{array}{l}{\frac{5 \times 10^{-19}}{(2-x)^{2}}=\frac{20 \times 10^{-19}}{x^{2}}} \\ \\{\frac{5}{(2-x)^{2}}=\frac{20}{x^{2}}} \\ \\{\frac{1}{(2-x)^{2}}=\frac{4}{x^{2}}} \\ \\{\text { Squaring both sides, } \frac{1}{(2-x)}=\frac{2}{x}}\end{array}
x = 4 – 2x
x + 2 x = 4
3x =4
x=\frac{4}{3}
x = 1.33 m.
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