Ask Questions, Get Answers

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

A non-conducting disc of radius a and uniform +ve surface charge density $\; \sigma\;$ is placed on the ground with it 's axis vertical . A particle of mass m and +ve charge q is dropped , along the axis of the disc from a height H with zero initial velocity . Find the value of H if the particle just reaches the disk .$\;\large\frac{q}{m}=4 \in_{0} g/\sigma$ . Find the position of particle at which the particle is in equilibrium. Find the energy of the particle at $\;H=\large\frac{a}{\sqrt{3}}\;$ assuming the gravitational energy at $\;H=0\;$ to be zero

$(a)\;mga\qquad(b)\;\sqrt{3} mga\qquad(c)\;\large\frac{mga}{\sqrt{3}}\qquad(d)\;None\;of\;these$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;\sqrt{3} mga$
Explanation :
Electrostatic energy at H=0 is:
$E_{e}=\large\frac{\sigma q a }{2 \in_{0}}$
At $\;H=\large\frac{a}{\sqrt{3}}\;, \quad\; E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+H^2}-H]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+\large\frac{a^2}{3}}-\large\frac{a}{\sqrt{3}}]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}} \;[\large\frac{2a}{\sqrt{3}}-\large\frac{a}{\sqrt{3}}]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;\large\frac{a}{\sqrt{3}}$
$T.E=\large\frac{mga}{\sqrt{3}}+\large\frac{\sigma q a}{2 \in_{0} \sqrt{3}}$
$T.E=\sqrt{3} mg\;.$
answered Feb 19, 2014 by yamini.v

Related questions

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