Question

Metal sphere $A$in FIGURE has $4$units of negative charge and metal sphere $B$has $2$units of positive charge. The two spheres are brought into contact. What is the final charge state of each sphere? Explain.

Short answer

Each sphere has one unit of negative charge$-e$

Step-by-step solution

Sphere A

The unit of charge is charge $e$, which is equal to the negative charge of an electron and the positive charge of a proton. Because sphere $A$has four units of negative charge, its charge is

$$\begin{gathered} q_A=4(-e) \\ =-4e \end{gathered}$$

Sphere B

Because sphere $B$contains two units of positive charge, its charge is

$$\begin{gathered} q_B=2(+e) \\ =-2e \end{gathered}$$

When two sphere are in contact.

Charges transfer from sphere A to sphere B when the two spheres come into contact due to charge conservation. As a result, each sphere's final charge is

$$\begin{gathered} q_{A,\mathrm{f}}=q_{B,\mathrm{f}} \\ =\frac{-4e+2e}{2} \\ =-e \end{gathered}$$

Because the remaining charges are split between both spheres, we divide by two. As a result, each sphere contains a single unit of negative charge.