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As shown in the figure, charges `+q and -q` are placed at the vertices `B` and `C` of an isoscles triangle. The potential at the vertex `A` is A. `(1)/(4pi epsilon_(0)).(2q)/(sqrt(a^(2) + b^(2))`B. ZeroC. `(1)/(4pi epsilon_(0)).(q)/(sqrt(a^(2) + b^(2))`D. `(1)/(4pi epsilon_(0)).((-q))/(sqrt(a^(2) + b^(2))` |
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Answer» Correct Answer - B Potential at `A =` Potential due to `(+q)` charge `+`Potential due to `(-q)` charge `= (1)/(4pi epsilon_(0)). (q)/(sqrt(a^(2) + b^(2))) + (1)/(4pi epsilon_(0)) ((-q))/(sqrt(a^(2) + b^(2))) = 0` |
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