logo

Ask Questions, Get Answers

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

Consider a spherical surface of radius 4m centred at the origin . Point +q and -2q are fixed at points $\;A (2m , 0 , 0)\;$ and $\;B (8m , 0 ,0)\;$ respectively . The potential on the surface of sphere is

$(a)\;\large\frac{k\;q}{2}\qquad(b)\;-\large\frac{2\;k\;q}{4}\qquad(c)\;-\large\frac{k\;q}{2}\qquad(d)\;zero$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) zero
Explanation :
Let $\;P (x , y ,z)\;$ be any point on the sphere . From the property of the sphere
$x^2+y^2+z^2=4^2=16$
Further ,
$PA=\sqrt{(x-2)^2+y^2+z^2}$
$=\sqrt{x^2+y^2+z^2+4-4x}$
$=\sqrt{16+4-4x}$
$PA=\sqrt{20-4x}$
and $\;PB= \sqrt{(x-8)^2+y^2+z^2}$
$=\sqrt{x^2+y^2+z^2+64-16x}$
$=\sqrt{80-16x}$
$=2 \sqrt{20-4x}$
Let $\;V_{P}\;$ be potential at P
$V_{P}=\large\frac{1}{4 \pi \in_{0}}\;[\large\frac{q}{PA}-\large\frac{2\;q}{PB}]$
$=\large\frac{1}{4 \pi \in_{0}}\;[\large\frac{q}{\sqrt{20-4x}}-\large\frac{2\;q}{2 \sqrt{20-4x}}]$
$=0\;.$
answered Feb 12, 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
...