logo

Ask Questions, Get Answers

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

Keeping the voltage of the charging source constant, what would be the percentage change in the energy stored in a parallel plate capacitor, if the separation between its plates were to be decreased by $10 \% $?

Can you answer this question?
 
 

1 Answer

0 votes
Energy stored in a parallel plate capacitor is $ U=\large\frac{1}{2}$$CV^2$
$\qquad =\large\frac{1}{2} \frac{\in_0 A}{d}$$V^2$------(i)
When separation between the plates is decreased by 10%, $d'=d \large\frac{10}{100}$$d=0.9 d$
Change in energy $(U' - U)=\large\frac{1}{2} \in_0 $$AV^2 \bigg[ \large\frac{1}{0.9} -\frac{1}{1}\bigg]$
% age change in energy= $\large\frac{(U'-U)}{U} $$ \times 100 =\large\frac{100}{9} $$=11.1 \%$
answered Jun 19, 2014 by meena.p
 

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
...