Answer to Question #118564 in Electricity and Magnetism for Pieter Coetze

Question

Two spheres, P and Q with charges of +4 microcoulomb and -6 microcoulomb are brought into contact and separated again.
What is the charge on each sphere after contact?

Answer 1

The potential of the first sphere before after contact is:

$$\varphi_1' = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1'}{R_1}$$

where $q_1'$ is the charge on the first sphere after contact and $R_1$ is the radius of the first sphere.

The potential of the second sphere after contact is:

$$\varphi_2' = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_2'}{R_2}$$

where $q_2'$ is the charge on the second sphere after contact and $R_2$ is the radius of the second sphere.

As far as the charge conserves, the total anoum of charge will not change:

$$q_1' + q_2' = 4\mu C + (-6\mu C) = -2\mu C$$

After contact the potentials of the spheres will be equal to each other:

$$\varphi_1' = \varphi_2' \Rightarrow \dfrac{q_1'}{R_1} = \dfrac{q_2'}{R_2}$$

Express $q_1'$ and substitute it to the previous equation:

$$q_1' = q_2'\dfrac{R_1}{R_2}\\ q_2'\dfrac{R_1}{R_2} + q_2' = -2\mu C\\ q_2' = -2 \dfrac{R_2}{R_1 + R_2} \mu C$$

The charge of the first sphere will be:

$$q_1' = -2 \dfrac{R_1}{R_1 + R_2} \mu C$$