logo

Ask Questions, Get Answers

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

A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius b $\;(b >a)\;$ . The solid sphere is given a charge Q . Potential at the surface of the solid sphere is $\;V_{a}\;$ and at the surface of spherical shell is $\;V_{b}\;$. If we put -4Q charge on the shell the potential difference between the sphere and shell becomes $\;( \bigtriangleup V )\;.$ Now we earthed the outer shell and the charge on the ouer shell is $\;q_{1}\;$.Now we removed the earthing connection from the shell and earthed the inner solid sphere. Charge on solid sphere becomes $\;q_{2}\;$. Answer the following question based on the passage. Potential difference $\;(\bigtriangleup V) \;$ measured between the inner solid sphere and outer shell after putting a charge -4Q is

$(a)\;V_{a}-3V_{b}\qquad(b)\;3(V_{a}-V_{b})\qquad(c)\;V_{a}\qquad(d)\;V_{a}-V_{b}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) \;$V_{a}-V_{b}$
Explanation :
$V_{a}=\large\frac{kQ}{a}$
$V_{b}=\large\frac{kQ}{b}$
After putting charge on shell
$V_{a}^{|}=\large\frac{kQ}{a}-\large\frac{4kQ}{b}$
$V_{b}^{|}=\large\frac{kQ}{b}-\large\frac{4kQ}{b}$
$V_{a}^{|}-V_{b}^{|}=\large\frac{kQ}{a}-\large\frac{kQ}{b}=V_{a}-V_{b}$
Hence $\; \bigtriangleup V =V_{a}-V_{b}\;.$
answered Feb 20, 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
...