logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A charged +Q is uniformly distributed in a spherical volume of radius R . A particle of charge +q and mass m projected with velocity $\;v_{0}\;$ from the surface of the spherical volume to its centre inside a smooth tunnel dug across the sphere . The minimum value of $\;v_{0}\;$ such that it just reaches the centre of the spherical volume is

$(a)\;\sqrt{\large\frac{kQq}{2mR}}\qquad(b)\;\sqrt{\large\frac{kQq}{4mR}}\qquad(c)\;\sqrt{\large\frac{2kQq}{4mR}}\qquad(d)\;\sqrt{\large\frac{kQq}{mR}}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) $\;\sqrt{\large\frac{kQq}{mR}}$
Explanation :
Work done by electric field on particle = Charge in kinetic energy of particle
Electric field at a distance r from the centre is E = $\;\large\frac{kQr}{R^3}$
Work done by electric field from r to rdr is
$dW=qE dr$
$W=\int_{R}^{0}\;qE\; dr$
$W=\large\frac{kqQ}{R^3}\;\int_{R}^{0}\;r\;dr$
$W=-\large\frac{kqQ}{2R^3} \;R^2=-\large\frac{kqQ}{2R}$
For $\;v_{0}\;$ to be minimum velocity on surface velocity at centre should be zero
Therefore
$\bigtriangleup K.E=\large\frac{1}{2} m (0)^2 - \large\frac{1}{2} mv_{0}^2=-\large\frac{1}{2} mv_{0}^2$
$W=\bigtriangleup K.E$
$-\large\frac{kqQ}{2R}=-\large\frac{1}{2}\;mv_{0}^2$
$v_{0}=\sqrt{\large\frac{kqQ}{mR}}$
answered Feb 13, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...