Question

A 4.8 μC point charge is placed in an external uniform electric field that has a magnitude of 72000 N/C. At what distance from the charge is the net electric field zero?

Answer 1

The electric field due to charge q is given by

$E = \frac{k|q|}{r^2}$

The distance when the electric field is zero is given by

$E_1 = E_2$

$E_1 = \frac{kq}{r^2}$

$72000\; N/C= \frac{(9 \times 10^9)(4.8\times 10^{-6}C)}{r^2}$

$r^2 = 6$

$r =\sqrt6=2.45\; m$

Answer: 2.45 m