Question

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

Answer 1

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.