Question

a 5.2nC charge exerts a repulsive force of 30mN on a second charge which is located at a distance of 60m away from it. what is the charge (magnitude and sign) of the second charge?

Answer 1

Answer:

Charge, $q_2=2.3\ C$

Explanation:

It is given that,

Charge 1, $q_1=5.2\ nC=5.2\times 10^{-9}\ C$

Repulsive force, $F=30\ mN=30\times 10^{-3}\ N$

Distance between two charges, d = 60 m

We need to find the magnitude of second charge. The electric force between two charge is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$q_2$ is second charge

$q_2=\dfrac{Fd^2}{kq_1}$

k is electrostatic constant

$q_2=\dfrac{30\times 10^{-3}\times (60)^2}{9\times 10^9\times 5.2\times 10^{-9}} \\\\q_2=2.3\ C$

So, the magnitude of second charge is 2.3 C. Hence, this is the required solution.