Question

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 F capacitor at a potential difference of 10000 V will be:

• A. 1/4
• B. 1
• C. 1/2
• D. 2

Answer 1

Answer: (B)

1

When a Force FN is applied the potential energy is given by $U=\frac{1}{2}Fx$ Energy stored by the capacitor$=\frac{1}{2}C{V}^2$

$\therefore Ratiois=\frac{\frac{1}{2}Fx}{\frac{1}{2}C{V}^2}=\frac{5000\times 0.2}{10\times {10}^{-6}\times ({10}^4{)}^2}=1$