Question

A charge q is placed at the center of the line joining two equal charges Q. The system of three charges will be in equilibrium if q is equal to

• A. $\frac{-Q}{2}$
• B. $\frac{-Q}{4}$
• C. $\frac{+Q}{2}$
• D. $\frac{+Q}{4}$

Answer 1

The correct option is B $\frac{-Q}{4}$

Step 1,

According to the question, q is placed between the two charges q. Given situation can be represented as

Step 2,

Now,

For system to be in equilibrium

$F_{BA}+F_{BC}=0$

$F_{BA}=-F_{BC}$

$\frac{K{Q}^2}{(2x{)}^2}=-\frac{KQq}{x^2}$

$q=-\frac{Q}{4}$

Hence for the system to be in equilibrium condition should be $q=-\frac{Q}{4}$