Question

Two point charges of $20 μC$ and $80 μC$ are 10 cm apart. Where will the electric field strength be zero on the line joining the charges ?

Easy

Solution

Let the electric field strength will be zero at a distance x. Then
$\frac{1}{4 {πε}_{0}} \times \frac{(20 \times 10^{- 6})}{x^{2}} = \frac{1}{4 {πε}_{0}} \times \frac{(80 \times 10^{- 6})}{(10 - x)^{2}} or \frac{(10 - x)^{2}}{x^{2}} = 4 or \frac{10 - x}{x} = \pm 2$
This gives x = - 10 cm or 3 . 33 cm i.e., 0.033 m

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