JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 & 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.…
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