Answer to Question #246818 in Electricity and Magnetism for Zee

Question

A metal sphere with a negative charge of 3.00 µC is placed 12.0 cm from another similar metal sphere with a positive charge of 2.00 µC. The two spheres momentarily touch, and then return to their original positions. Calculate the electrostatic force acting on the two metal spheres. 

Answer 1

As the two spheres are identical then after contact charge in them will be equal and will be half of the total charge.

Charge of the first sphere is

$Q_1=-3µC$

charge of the second sphere is

$Q_2=2µC$

After contact charge in each sphere is

$Q=\frac{Q_1+Q_2}{2} \\ Q=\frac{-3µ C+2µC}{2}=-0.5µ C=-0.5\times 10^{-6}C$

Distance between them is

d=12cm =0.12 m

So now electrostatic force between them is

$F=k\frac{Q^2}{d^2} \\ Wherek=9\times 10^9Nm^2/C^2 \\ \therefore F=9\times 10^9\times \frac{(0.5\times10^{-6})^2}{(0.12)^2} \\ F=1.56\times 10^{-1}N$

As nature of the two charged sphere are same.

Therefore the force between them is repulsive