Question

20 positively charged particles are kept fixed on the x-axis at point x=1m,2m,3m,.......,20m.the first particle has a charge 1.0×10^-8c,the second 8×10^-6c,the third 27×10^-6c and so on.find the magnitude of the electric force acting on a 1c charge placed at origin.

Answer 1

The magnitude of electric force acting on charge $Q_2$ because of charge $Q_1$ separated by a distance $r$ is

$F_{12}=\frac{1}{4\pi \epsilon_0} \frac{Q_1Q_2}{r^2}$.

Here the total force acting on 1 $\mu$C charge is

[tex]F=\sum _{n=1}^{20}\frac{10^{-12}}{4\pi \epsilon_0} \frac{n^3}{n^2}\\ F=\sum _{n=1}^{20}\frac{10^{-12}}{4\pi \epsilon_0} n\\ F=\frac{10^{-12}}{4\pi \epsilon_0}\sum _{n=1}^{20} n\\ F=\frac{10^{-12}}{4\pi \epsilon_0}\frac{20(20+1)}{2} \\[/tex]

[tex]F=10^{-12}*9*10^9*210\\ F=1.89 \;N[/tex]