InterviewSolution
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InterviewSolution
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Two point charge `Q_(a)` and `Q_(b)` are positional at point A and B. The field strength to the right of charge `Q_(b)` on the line that passes through the two charges varies according to a law represented schematically in fig. (without employing a definite scale). The field strength is assumed to be positive if its direction coincides with the positive direction of the x-axis. The distance between the charges is `l=21 cm`. (a) Find the sign of the charges. (b) Find the ration between the absolute value of charge `Q_(a)` and `Q_(b)`. (c ) Find the coordinate x of the point where the field strength is maximum. |
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Answer» Correct Answer - A::B::C Over charge `Q_2` field intensity is infinite along negatve `x` -axis. Therefore `Q_2` is negative. Beyon `xgt(l+a)`, field intensity is positive. Therefore `Q_1` is positive b. At `x=l+a`, field intensity is zero. `:. (kQ_1)/((l+a)^2)=(kQ_2)/a^2` or `|Q_1/Q_2|=((l+a)/a)^2` c. Intensity at distance `x` from charge `2` would be `E=(kQ_1)/((x+l)^2-(kQ_2)/x^2` For `E` to be maximum `(dE)/(dx)=0` or `-(2kQ_1)/((x+l)^3)+(2kQ_2)/x^3=0` or `(1+l/x)^3=Q_1/Q_2=((l+a)/a)^2` or `1+l/x=((l+a)/a)^(2//3))` or 1+l/x=((l+a)/a)^(2//3)` or `x=l/(((l+a)/a)^(2//3)-1)` or `b=l/(((l+1)/a)^(2//3)-1)` |
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