Q.
Four charges are placed on corners of a square as shown in figure having side of $5\ cm.$ If $Q$ is one microcoulomb, then electric field intensity at centre will be
A) $1.02 \times 10^{7}N/C$ upwards
B) $2.04 \times 10^{7}N/C$ downwards
C) $2.04 \times 10^{7}N/C$ upwards
D) $1.02 \times 10^{7}N/C$ downwards
Explanation
(a) Side $a = 5 \times 10^{- 2}m$
Half of the diagonal of the square $r = \frac{a}{\sqrt{2}}$
Electric field at centre due to charge $Q\ E = \ \frac{kQ}{\left( \frac{a}{\sqrt{2}} \right)^{2}}$
Now field at $O = \sqrt{E^{2} + E^{2}} = E\sqrt{2} = \frac{kq}{\left( \frac{a}{\sqrt{2}} \right)^{2}}.\sqrt{2}$
$= \frac{9 \times 10^{9} \times 10^{- 6} \times \sqrt{2} \times 2}{\left( 5 \times 10^{- 2} \right)^{2}} = 1.02 \times 10^{7}N/C$ [upward]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.