Question

Two charges of value $2\mu C$ and $-50\mu C$ are placed 80 cm apart. Calculate the distance of the point from the smaller charge where the intensity is zero.

Answer 1

Amount of first charge is $=2\times {10}^{-6}C$

Amount of second charge is $=50\times {10}^{-6}C$

Distance $=80cm$

Electric Field $=\frac{k{q}^1{q}^2}{r}$

Electric Field will be zero

E.F.$=\frac{0(K{x}^2\times {10}^{-}6\times 50\times {10}^{-}6)}{80}+\left(\frac{k\times 2\times {10}^{-6}}{r}\right)$

$=0\left(\frac{{10}^{-4}}{80}\right)+(2\times \frac{{10}^{-6}}{r})$

$=0\left(\frac{{10}^{-4}}{80}\right)$

$=-(2\times \frac{{10}^{-6}}{r})\left(\frac{{10}^2}{80}\right)$

$=\frac{2}{r\left(\frac{10}{8}\right)}$

$=\frac{2}{r\left(\frac{10}{4}\right)}$

$=\frac{1}{r}$

$R=0.4$