Question

Two balls of radius 2.00 mm have a separation between their centers of 5.33 cm. The same electric charge is placed on both balls so that there is a repulsive force between them of 2.75 N. Assume that the charge is uniformly distributed over each ball. What is the magnitude of the charge

Answer 1

Answer:

Charges, $q_1=q_2=4.03\ \mu C$  

Explanation:

Given that,

Radius of the balls is 2 mm

The force of repulsion acting on the balls, F = 2.75 N

The separation between the spheres, d = 5.33 cm = 0.0533 m

The force of repulsion between charges is given by :

$F=\dfrac{kq^2}{d}$

$q^2=\dfrac{Fd}{k}$

$q^2=\dfrac{2.75 \times 0.0533 }{9\times 10^9}$

$q^2=1.62\times 10^{-111}\ C$

$q=4.03\times 10^{-6}\ C$

$q=4.03\ \mu C$

So, the charge on the spheres is $q=4.03\ \mu C$. Hence, this is the required solution.