InterviewSolution
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InterviewSolution
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A disc of radius `a//4` having a uniformly distributed charge `6C` is placed in the x-y plane with its centre at `(-a//2, 0, 0)`. A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from `x=a//4` to `x=5a//4`. Two point charges `-7C` and `3C` are placed at `(a//4, -a//4, 0)` and `(-3a//4, 3a//4, 0)`, respectively. Conisder a cubical surface formed by isx surfaces `x=+-a//2`, `y=+-a//2`, `z=+-a//2`. The electric flux through this cubical surface is A. `-2 C//in_(0)`B. `2 C//in_(0)`C. `10 C//in_(0)`D. `12 C//in_(0)` |
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Answer» Correct Answer - A Now charge on disc contributing to electric flux through cubical surface `= (6C)/(2) = 3C` Charge on rod contributing to electric flux through cubical surface `= (8C)/(4) = 2C` `-7 C` charge also contributes to electric flux but `3C` charge si outside the cubical surface,so it does not contribute to electric flux. Therefore `q_(enclosed) = 3C +2C - 7C = -2C` Hence electrc flux =, `phi_(C) = (q_(enclosed))/(in_(0)) = (-2C)/(in_(0))` |
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