SATHEE: Electrostatics Question 819

Electrostatics Question 819

Question: A unit positive point charge of mass m is projected with a velocity V inside the tunnel as tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere (charge density $ρ$ ). 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

Options:

A) ${[ρ R^{2} / 4 m ε_{0}]}^{1 / 2}$

B) ${[ρ R^{2} / 24 m ε_{0}]}^{1 / 2}$

C) ${[ρ R^{2} / 6 m ε_{0}]}^{1 / 2}$

D) zero because the initial and the final points are at some potential

Show Answer

Answer:

Correct Answer: A

Solution:

[a] If we thraw the charged particle just right the tunnel. Hence, applying conservation of momentum between start point and center of tunnel we get $Δ K + Δ U = 0$

$or (0 + \frac{1}{2} m v^{2}) + q (V_{f} - V_{i}) = 0$

$or V_{f} = \frac{V_{s}}{2} (3 - \frac{r^{2}}{R^{2}}) = \frac{ρ R^{2}}{6 ε_{0}} (3 - \frac{r^{2}}{R^{2}})$

$Hence r = \frac{R}{2}$

$V_{f} = \frac{ρ R^{2}}{6 ε_{0}} (3 - \frac{R^{2}}{4 R^{2}}) = \frac{11 ρ R^{2}}{24 ε_{0}}; V_{i} = \frac{ρ R^{2}}{3 ε_{0}}$

$\frac{1}{2} m v^{2} = 1 [\frac{11 ρ R^{2}}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}]$

$= \frac{ρ R^{2}}{ε_{0}} [\frac{11}{24} - \frac{1}{3}] = \frac{ρ R^{2}}{3 ε_{0}} or V = {(\frac{ρ R^{2}}{4 m ε_{0}})}^{1 / 2}$ Hence velocity should be slightly greater than V.

Electrostatics Question 819

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Electrostatics Question 819

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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