Question

Two identical $+ve$ charges are at the ends of a straight line AB. Another identical $+ve$ charge is placed at ‘C’ such that $AB=BC$. (A, B and C being on the same line). Now the force on ‘A’ will :

• A. increase
• B. decrease
• C. remain same
• D. cannot be predicted

Answer 1

The correct option is A increase

Initially (when C was not present) the force on A $=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

Now when C is added, the net focre on A $=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}+\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{(2r{)}^2}$

(since $q_1=q_2=q$)

$=\frac{q^2}{4\pi {ϵ}_0}(\frac{1}{r^2}+\frac{1}{4r^2})$

$F=\frac{5q^2}{4\times 16\pi {ϵ}_0{r}^2}$

$\therefore$ F is increased by $\frac{1}{4\pi {ϵ}_0}\frac{q^2}{\left(4r^2\right)}$