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If 8 dipoles of charges of magnitude ± e are placed inside the cube. The electric flux coming out of cube will be(a) 16e/E0(b) 8e/E0(c) e/E0(d) zero |
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Answer» Answer is : (d) zero From Gauss theorem, \(\phi_E=\frac{q_{in}}{E_0}\) \(=\frac{+8e-8e}{E_0}=0\) |
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