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Three point charges are placed at the following points on x-axis : `3 muC` at `x = 0, -4 muC ` at `x = 50 cm` and `-5 muC` at `x = 50 cm` and `-5 muC` at `x = 120 cm`, Calculate the force on `-4 muC` charge. |
Answer» Here, force of repulsion `(F_(1)) on -4 muC` charge by `-5 muC` charge will be towards left, Fig `F_(1) = (1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2))` ` = (9xx10^(9)xx4xx10^(-6)xx5xx10^(-6))/((0.7)^(2)) = 0.367 N` And force of attraction by` + 3` muC charge on `-4 muC` charge will also be towards left. `F^(2) = ((9xx10^(9)xx4xx10^(-6)xx3xx10^(-6))/((0.5)^(2))) = 0.432 N` Total force,` F = F_(1) + F_(2)` `0.367 + 0.432 = 0.799 N` towards left. |
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