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

The Correct Option is 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. 
\(\vec{ E }=\frac{ kq \vec{ r }}{ r ^{3}}=\frac{1(-2 Q ) \frac{ a }{2}(\hat{ x }+\hat{ y }+\hat{ z })}{4 \pi \varepsilon_{0}\left(\frac{ a }{2} \sqrt{3}\right)^{3}}\)

\(\vec{ E }=\frac{-2 Q (\hat{ x }+\hat{ y }+\hat{ z })}{3 \sqrt{3} \pi a ^{2} \varepsilon_{0}}\)

Hence, The correct answer is option (B): \(\frac{-2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })\)