Q. Charges of $+ \frac{10}{3} \times 10^{- 9}\ C$ are placed at each of the four corners of a square of side 8$cm$. The potential at the intersection of the diagonals is
A) $150\sqrt{2}\ volt$
B) $1500\sqrt{2}\ volt$
C) $900\sqrt{2}\ volt$
D) $900\ volt$
Explanation
(b) Potential at the centre $O,\ V = 4 \times \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{a/\sqrt{2}}$
[image:1_image309.wmf]Where $Q = \frac{10}{3} \times 10^{- 9}C$ and $a = 8cm = 8 \times 10^{- 2}m$
So $V = 4 \times 9 \times 10^{9} \times \frac{\frac{10}{3} \times 10^{- 9}}{\frac{8 \times 10^{- 2}}{\sqrt{2}}} = 1500\sqrt{2}\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.