Question

A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is : 

• A. $-\frac{Q}{3\sqrt{3} \pi \epsilon_0 a^2} (\hat{i} + \hat{j} + \hat{k})$
• B. $\frac{Q}{3\sqrt{3} \pi \epsilon_0 a^2} (\hat{i} + \hat{j} + \hat{k})$
• C. $-\frac{2Q}{3\sqrt{3} \pi \epsilon_0 a^2} (\hat{i} + \hat{j} + \hat{k})$
• D. $\frac{2Q}{3\sqrt{3} \pi \epsilon_0 a^2} (\hat{i} + \hat{j} + \hat{k})$

Hint:

Use the superposition principle: a missing charge $+Q$ or a replaced charge $-Q$ can be treated as adding a negative charge to a perfectly symmetric system.

Answer 1

The Correct Option is D

Step 1: If all vertices had $+Q$, the field at the center would be zero by symmetry.
Step 2: Having $-Q$ at the origin is equivalent to having $+Q$ there and adding a compensating charge of $-2Q$.
Step 3: Thus, the net field is that of a single $-2Q$ charge placed at the origin.
Step 4: Center $P = (a/2, a/2, a/2)$. Distance $r = \sqrt{(a/2)^2 \times 3} = \frac{\sqrt{3}a}{2}$.
Step 5: $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{(-2Q)}{r^2} (-\hat{r})$. Direction is from $P$ to origin: $-(\frac{a}{2}\hat{i} + \frac{a}{2}\hat{j} + \frac{a}{2}\hat{k})/r$.
Step 6: Resulting field points from origin to center: $\frac{2Q}{4\pi\epsilon_0 (\frac{3a^2}{4})} \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} = \frac{2Q}{3\sqrt{3}\pi\epsilon_0 a^2}(\hat{i}+\hat{j}+\hat{k})$.