Question

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

Answer 1

Given $q_1=5\times {10}^{-19}C$ and $q_2=20\times {10}^{-19}C$ with $r=2m$

Suppose at a point $x$ meter away from q_1 the field is zero then at that point the field due to individual charges should be equal and opposite

$\frac{q_1}{x^2}=\frac{q_2}{(2-x{)}^2}$ which lead us to $x=2/3$ meter from charge $5\times {10}^{-19}C$