JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{-2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})\)
Step-by-step Solution
Detailed explanation
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.
\(\overrightarrow{ E }=\frac{ kq \overrightarrow{ r }}{ r ^{3}}=\frac{1(-2 Q ) \frac{ a }{2}(\hat{ x }+\hat{ y }+\hat{ z })}{4 \pi \varepsilon_{0}\left(\frac{ a }{2} \sqrt{3}\right)^{3}}\)
\(\overrightarrow{ E }=\frac{-2 Q (\hat{ x }+\hat{ y }+\hat{ z })}{3 \sqrt{3} \pi a ^{2} \varepsilon_{0}}\)
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