Question

Eight equal charges each +Qare kept at the corners of a cube. Net electric field at the centre will be$\left( k = \frac{1}{4\pi\varepsilon_{0}} \right)$

• A. $\frac{kQ}{r^{2}}$
• B. $\frac{8kQ}{r^{2}}$
• C. $\frac{2kQ}{r^{2}}$
• D. Zero

Answer 1

Answer: (C)

$\frac{2kQ}{r^{2}}$

Answer 2

For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The total flux $\varphi_{T} = \frac{Q}{\varepsilon_{0}}$. Therefore the flux through one cube will be $\varphi_{cube} = \frac{Q}{8\varepsilon_{0}}.$ The cube has six faces and flux linked with three faces (through A) is zero, so flux linked with remaining three faces will $\frac{\varphi}{8\varepsilon_{0}}.$ Now as the remaining three are identical so flux linked with each of the three faces will be $= \frac{1}{3} \times \left\lbrack \frac{1}{8}\left( \frac{Q}{\varepsilon_{0}} \right) \right\rbrack = \frac{1}{24}\frac{Q}{\varepsilon_{0}}$.