Question

10. Three charges Q, +q and + q are placed at the vertices of an equilateral triangle of side l as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to

• A. -\frac{q}{2}
• B. -q
• C.
  • q
• D. zero

Answer 1

Answer: (A)

-\frac{q}{2}

Answer 2

The electrostatic energy of the system is given by

$$U = \frac{k}{l} \Big[Q(+q) + Q(+q) + (+q)(+q)\Big] = \frac{k}{l} (2Qq + q^2)$$

Setting $U=0$ we get

$$2Qq + q^2 = 0 \quad \Rightarrow \quad q(2Q + q)=0$$

Since $q \neq 0$,

$$2Q+q=0 \quad \Rightarrow \quad Q=-\frac{q}{2}$$