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Two charge 5xx10^(-8)Cand -3xx10^(-8) Care located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. |
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Answer» Solution :Here are two possibilities. First possibillty : Let charge `q_(1) = 5xx10^(-8) ` C and `q_(2) =-3XX10^(-8) ` C are placed at point A and B respectively. AB = 0.16 m  Let electric potential at point P is zero `:.` Electric potential at Point P `V = V_(1) +V_(2)` `O=(kq_(1))/(x) +(kq_(2))/(0.16-x)` `O = (5xx10^(-8))/(x) -(3xx10^(-8))/(0.16-x) [ because ` Dividing by k] `:. (3xx10^(-8))/(0.16-x)=(5xx10^(-8))/(x)` `:. (3)/(0.16-x) =(5)/(x)` `:. 3x=0.8 -5x` `:. 8x=0.8` `:. x = 0.1 `m `:. ` It is zero from 0.1 m away from `5xx10^(-8) ` C or It is zero from 0.06 m away from `-3 xx10^(-8) C ` Second possibillity :  `V= V_(1) + V_(2)` `:. 0 =(kq_(1))/(x) (kq_(2))/(x-16)` `:. (q_(1))/(x) = (q_(2))/(x-16)` `:. (5)/(x XX 10^(-2))= (3)/((x-16)xx10^(-2))` `:. (5)/(x) =(3)/(x-16)` `:. 5x-80=3x` `:. 2x=80 ` `:. x = 40 cm` `:.` 40 cm away from `q_(1)` cahrge or 24 cm away from `q_(2)` charge. |
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