Question

Three point charges of +2 $\mu$C, -3 $\mu$C, and -3 $\mu$C are kept at the vertices A, B and C, respectively, of an equilateral triangle of side 30 cm. What should be the sign and magnitude of the charge (q) to be placed at the midpoint (M) of side BC so that the charges at A remains in equilibrium?

Answer 1

Net electric field at A due to charges at B and C is

$E_A=2E_{Ac}sin{50}^o=2\times 9\times {10}^9\times \frac{3\times {10}^{-6}}{(0.20{)}^2}\times \frac{\sqrt{3}}{2}$

$=\sqrt{3}\times 6.75\times {10}^5$

$AM=\sqrt{(20{)}^2-(10{)}^2}=\sqrt{400-100}=10\sqrt{3}cm$

Let the charge at M be q. Charge q should be positive so that there can be repulsion between the charges at A and M.

$E_{AM}=9\times {10}^9\times \frac{q}{(10\sqrt{3}/100{)}^2}=3q\times {10}^{11}$

For A to be in equilibrium

$E_A=E_{AM}\Rightarrow \sqrt{3}\times 6.75\times {10}^5=3q\times {10}^{11}$

or $q=\frac{9\sqrt{3}}{4}\times {10}^{-6}=\frac{9\sqrt{3}}{4}\mu C$