Question

A ball at O is in equilibrium as it is attached with two strings AO and DO which are tied at A and D. $AO=DO=a\sqrt{5}$.

The charges at A, B, C and D are +q, +Q, +2Q and -q respectively. The ball O is positively charged then

• A. The ball O can not remain in equilibrium.
• B. If the charge at C is +Q, the ball will remain in equilibrium.
• C. The ball at O will remain in equilibrium.
• D. If the charges at A and D and charges at B and C are interchanged, the ball will remain in equilibrium.

Answer 1

Answer: (C), (D)

The correct options are
C The ball at O will remain in equilibrium.
D If the charges at A and D and charges at B and C are interchanged, the ball will remain in equilibrium.

For ball O (positively charged) to be in equilibrium at the middle of BC.
$[\frac{q}{4\pi {ϵ}_0(a\sqrt{5}{)}^2}+\frac{q}{4\pi {ϵ}_0(a\sqrt{5}{)}^2}]cos\theta +\frac{Q}{4\pi {ϵ}_0{a}^2}=\frac{2Q}{4\pi {ϵ}_0{a}^2}$
$\frac{2qcos\theta }{5}+Q=2Q\Rightarrow q=\frac{5\sqrt{5}Q}{2}$

If $q=\frac{5\sqrt{5}Q}{2}$, ball will be in equilibrium even if the charges are interchanged.
If the charge at C is +Q, then
$\frac{2qcos\theta }{5}+Q=Q$
$\frac{2qcos\theta }{5}=0,q\ne 0,\theta ={90}^{\circ }$