A small point charge ( + q = 1 μ C , m = 3 kg ) is started from a very long distance towards a fixed point charge ( + Q = 1 mC ) . The initial line of motion of q is at a distance of d = 1 m from the

Question

A small point charge ( + q = 1 μ C , m = 3 kg ) is started from a very long distance towards a fixed point charge ( + Q = 1 mC ) . The initial line of motion of q is at a distance of d = 1 m from the line passing from the centre of Q , (as shown in figure). Take V 0 ​ = 2 m / s . Determine the least distance between the two point charges (in m)

TestHub · Accepted Answer

Due to repulsion forces the point of q will be a line curved away from Q .

at least distance, rl we can conserve angular momentum \& energy of system. $mv_{0} d = mvr$ $&0 + \frac{1}{2} m v_{0}^{2} = \frac{KqQ}{r} + \frac{1}{2} m v^{2}$ $\frac{1}{2} m v_{0}^{2} = \frac{K qQ}{r} + \frac{1}{2} m (\frac{v _{0} d}{r})^{2}$ $\frac{1}{2} m v_{0}^{2} (1 - \frac{d ^{2}}{r ^{2}}) = \frac{K qQ}{r}$ $\frac{1}{2} m v_{0}^{2} (r^{2} - d^{2}) = K qQ r$ $(\frac{1}{2} m v_{0}^{2}) r^{2} - (KqQ) r - \frac{1}{2} m v_{0}^{2} d^{2} = 0$ $r = \frac{K qQ + ( K qQ ) ^{2} + m ^{2} v _{0}^{4} d ^{2}}{m v _{0}^{2}} = 2 m$ $KqQ = 9$ $m v_{0}^{2} = 12$ $d = 1$

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