Question

When two spheres having $2$Q and $-$Q are placed at a certain distance, the force acting between them is F. Then they are connected by a conducting wire and again separated from each other. How much force will act between them if the separation now is the same as before?

• A. F
• B. $\frac{F}{2}$
• C. $\frac{F}{4}$
• D. $\frac{F}{8}$

Answer 1

Answer: (D)

$\frac{F}{8}$

According to Coulomb's law, force between two charge particle $q_1$ and $q_2$ at a distance r is $\frac{K{q}_1{q}_2}{r^2}$.

Given, Initial Force(magnitude) = $\frac{2K{Q}^2}{r^2}$ = F.

But, when they are connected by a wire, total charge i.e 2Q +(-Q) = Q will distribute equally (Q/2 each).

Now the force will be $\frac{K{Q}^2}{2\times 2\times r^2}$ = $\frac{K{Q}^2}{4r^2}$ =$\frac{F}{8}$.

Therefore, D is the correct option.