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. $\frac{- Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
• B. $\frac{-2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
• C. $\frac{2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
• D. $\frac{ Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$

Answer 1

Answer: (B)

We can replace $-Q$ charge at origin by $+ Q$ and $-2 Q .$
Now due to $+Q$ charge at every corner of cube. Electric field at center of cube is zero so now net electric field at center is only due to $-2 Q$ charge at origin.
$\vec{ E }=\frac{ kq \vec{ r }}{ r ^{3}}=\frac{1(-2 Q ) \frac{ a }{2}(\hat{ x }+\hat{ y }+\hat{ z })}{4 \pi \varepsilon_{0}\left(\frac{ a }{2} \sqrt{3}\right)^{3}}$
$\vec{ E }=\frac{-2 Q (\hat{ x }+\hat{ y }+\hat{ z })}{3 \sqrt{3} \pi a ^{2} \varepsilon_{0}}$