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Two point charges +8q and -2q are located at x = 0 and x = L respectively. The point on x axis at which net electric field is zero due to these charges is- (i) 8L (ii) 4L (iii) 2 L (iv) L |
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Answer» Answer is (iii) 2L Let P is the observation point at a distance r from -2q and at (L+r) from +8q. Given Now, Net EFI at P = 0 ∴ \(\overrightarrow{E_1}\) = EFI (Electric Field Intensity) at P due to +8q \(\overrightarrow{E_2}\) = EFI (Electric Field Intensity) at P due to -2q \(|\overrightarrow{E_1|}=|\overrightarrow{E_2}|\) ∴ \(\frac{K(8q)}{(L+r)^2}=\frac{1}{(r)^2}\) ∴ \(\frac{4}{(L+r)^2}=\frac{1}{(r)^2}\) 4r2 = (L+r)2 2r = L+r r = L ∴ P is at x = L + L = 2L from origin 2L |
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