First slide

Electrostatic potential energy

Question

A unit positive point charge of mass m is projected with a velocity V inside the tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere. The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel is equal to

Moderate

A

${[{ρR}^{2} / 4 {mε}_{0}]}^{1 / 2}$
B

${[{ρR}^{2} / 24 {mε}_{0}]}^{1 / 2}$
C

${[{ρR}^{2} / 6 {mε}_{0}]}^{1 / 2}$
D

zero because the initial and the final points are at same potential

Solution

If we throw the charged particle just right of the center of the tunnel, the particle will cross the tunnel. Hence, applying conservation of energy between starting point and center of tunnel we get
$ΔK + ΔU = 0$
or $(0 + \frac{1}{2} {mv}^{2}) + q (V_{f} - V_{i}) = 0$
or $V_{f} = \frac{V_{s}}{2} (3 - \frac{r^{2}}{R^{2}}) = \frac{{ρR}^{2}}{6 ε_{0}} (3 - \frac{r^{2}}{R^{2}})$
Here $r = \frac{R}{2}$
$V_{f} = \frac{{ρR}^{2}}{6 ε_{0}} (3 - \frac{R^{2}}{4 R^{2}}) = \frac{11 {ρR}^{2}}{24 ε_{0}}; V_{i} = \frac{{ρR}^{2}}{3 ε_{0}}$
$\frac{1}{2} {mv}^{2} = 1 [\frac{11 {ρR}^{2}}{24 ε_{0}} - \frac{{ρR}^{2}}{3 ε_{0}}] = \frac{{ρR}^{2}}{ε_{0}} [\frac{11}{24} - \frac{1}{3}] = \frac{{ρR}^{2}}{8 ε_{0}}$
or $V = {(\frac{{ρR}^{2}}{4 {mε}_{0}})}^{1 / 2}$
Hence velocity should be slightly greater than V.

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