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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

  1. A \(\frac{-Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})\)
  2. B \(\frac{-2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})\)
  3. C \(\frac{2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})\)
  4. D \(\frac{ Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })\)
Verified Solution

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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