Question

Equal charges of $\frac{10}{3} \times 10^{- 9}$ coulomb are lying on the corners of a square of side 8 cm. The electric potential at the point of intersection of the diagonals will be

• A. 900 V
• B. $900\sqrt{2}V$
• C. $150\sqrt{2}V$
• D. $1500\sqrt{2}V$

Answer 1

Answer: (D)

$1500\sqrt{2}V$

Answer 2

Potential at the centre O

$V = 4 \times \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{a/\sqrt{2}}$ given $Q = \frac{10}{3} \times 10^{- 9}C$

⇒ $a = 8cm = 8 \times 10^{- 2}m$

$V = 5 \times 9 \times 10^{9} \times \frac{\frac{10}{3} \times 10^{- 9}}{\frac{8 \times 10^{- 2}}{\sqrt{2}}} = 1500\sqrt{2}volt$