Question

An automobile spring extends 0.2m for 5000N load. The ratio of potential energy stored in the spring when It has been compressed by 0.2 m to the potential energy stored in a 10 microF Capacitor at a potential difference of 10000V will be

Answer 1

Convert microfarads to farads :

0.000001 × 10 = 0.00001 Farads

Potential energy = 0.5CV

Where C is charge in farads and V is Voltage in volts

P. E = 0.5 × 0.00001 × 10000 = 0.1Joules

P. E in the spring is :

Force × height moved :

5000 × 0.2 = 1000 Joules

The ratio of potential energy stored on the spring to potential energy stored in the capacitor is :

1000 / 0.1 = 10000 / 1

10000 : 1

Answer 2

Explanation:

$\huge10000 : 1$