Question

A 31 mC charge is positioned on the x axis at x + 8 cm. Where should a -70 mC charge be placed (on the positive X-axis) to produce a net electric field of zero at the origin?

Answer 1

Answer:r=12.02 cm

Explanation:

Given

$q_1=31 mC=31\times 10^{-3}C$

Placed at x=8 cm

$q_2=-70 mc=-70\times 10^{-3} C$

placed at a distance, suppose r

Electric field due to positive charge will be away from the origin and electric field due to negative charge is towards the origin

Thus net effect will be zero

$E=\frac{kq}{r^2}$

$E_1=\frac{9\times 10^{9}\times 31\times 10^{-3}}{0.08^2}$

$E_2=\frac{9\times 10^{9}\times 70\times 10^{-3}}{r^2}$

Equate $E_1=E_2$

$\frac{31}{8^2}=\frac{70}{r^2}$

$r=8\times \sqrt{\frac{70}{31}}$

r=12.02 cm