Question

Consider a regular cube with positive point charge $+Q$ in all corners except for one which has a negative point charge $-Q$. Let the distance from any corner to the center of the cube be $r$. What is the magnitude of electric field at point $P$, the center of the cube?

• A. $E=\frac{7kQ}{r^2}$
• B. $E=\frac{1kQ}{r^2}$
• C. $E=\frac{2kQ}{r^2}$
• D. $E=\frac{6kQ}{r^2}$

Answer 1

The correct option is C $E=\frac{2kQ}{r^2}$

Due to symmetry, electric field due to pairs of charges at opposite corners will cancel each other except the opposite charges $+Q$ and $-Q$
Hence, resultant field strength at $P$ is equal to
$E=2\times \frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}=\frac{2kQ}{r^2}$