question_answer

An automobile spring extends 0.2 m for 5000 N load. The ratio of potential energy stored in this spring when it has been compressed by 0.2 m to the potential energy stored in a \[10\,\mu F\] capacitor at a potential difference of 10000 V will be

A)  1/4                       

B)         1                            

C)  1/2                       

D)         2

Correct Answer: B

Solution :

When a force of F newton is applied the potential energy is given by  \[U=\frac{1}{2}Fx\] Energy stored by capacitor is \[\frac{1}{2}C{{V}^{2}}.\] \[\therefore \]Ratio is   \[\frac{\frac{1}{2}Fx}{\frac{1}{2}C{{V}^{2}}}=\frac{5000\times 0.2}{10\times {{10}^{-6}}{{({{10}^{4}})}^{2}}}=1\]