Question

An automobile spring extends $0.2\text{m}$ for $5000\text{N}$ load. The ratio of potential energy stored in the spring to the potential energy stored in a $10\mu \text{F}$ capacitor at a potential difference of $10,000\text{V}$ will be

• A. $\frac{1}{4}$
• B. $1$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is B $1$

For the Spring:

Force, $F=5000\text{N}$

Extension, $x=0.2\text{m}$

Force in a spring having extension $x$ and spring constant $k$ is given by

$F=kx$

$\Rightarrow 5000=k\times 0.2$

$\Rightarrow k=25000\text{N/m}$

So potential energy stored in the spring is,

$P{E}_S=\frac{1}{2}k{x}^2=\frac{1}{2}\times 25000\times {0.2}^2=500\text{J}$

For the Capacitor:

Capacitance,
$C=10\mu \text{F};V=10,000\text{V}$

So potential energy stored in the capacitor is $P{E}_C=\frac{1}{2}C{V}^2=\frac{1}{2}\times 10\times {10}^{-6}\times 10,{000}^2=500\text{J}$

Hence, ratio of energy stored in spring to that in capacitor is

$\frac{P{E}_S}{P{E}_C}=\frac{500}{500}=1$

Hence, option $\left(b\right)$ is correct.

Key Concept - $1$. Spring potential energy $=\frac{1}{2}k{x}^2$ $2$. Potential energy in a capacitor $=\frac{1}{2}C{V}^2$