Question

A point charge of 4 μС is located at x =-30 cm, and a second point charge of-10 pC is located at x-+4.0 cm. where should a third charge of +6.0 pC be placed so that the electric field at x = 0 is zero?

Answer 1

Answer:

it is to be placed at x = 3.1 cm

Explanation:

Electric field due to charge placed at x = -30 cm

so we have

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

$E_1 = \frac{9 \times 10^9(4\times 10^{-12})}{0.30^2}$

$E_1 = 0.4 N/C$ towards left

now electric field due to another charge placed at x = 4 cm

$E_2 = \frac{9 \times 10^9(10\times 10^{-12})}{0.04^2}$

$E_2 = 56.25 N/C$ towards Right

now we need electric field due to 6 pC must be equal to

$E = 56.25 - 0.4$

$E = 55.85 N/C$ towards Left

so we have

$55.85 = \frac{9 \times 10^9(6 \times 10^{-12})}{r^2}$

$r = 0.031 m$

so it is to be placed at x = 3.1 cm