Question

When two sphere having $5Q$ and $-Q$ charge are placed at a certain distance, the force acting between them is $F$. Now 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. $\frac{5}{4}F$
• C. $\frac{4}{5}F$
• D. $\frac{F}{8}$

Answer 1

The correct option is C $\frac{4}{5}F$

For the first figure, i.e. figure (1), $F=\frac{K\times 5Q\times Q}{r^2}=5\frac{K{Q}^2}{r^2}$

Now for the second figure, i.e. figure (2),

In the second figure the charge is distributed equally among both.

$\therefore F^{\prime }=\frac{K\times 2Q\times 2Q}{r^2}=\frac{4K{Q}^2}{r^2}\therefore F^{\prime }=\frac{4F}{5}$