Question

A 50 pF capacitor and a 200 pF capacitor are both charged to 4.2 kV. They are then disconnected from the voltage source and are connected together, positive plate to positive plate and negative plate to negative plate. Find the energy lost when the connections are made.

Answer 1

The energy lost is zero.

Step-by-step explanation:

Given that,

First capacitor = 50 pF

Second capacitor = 200 pF

Potential = 4.2 kV

We need to calculate the energy lost

Using formula of energy lost

`E = E_(initial)-E_(final)`

Put the value into the formula

`E=(1)/(2)C_(1)V^2+(1)/(2)C_(2)V^2-\frac{}{}(C_(1)+C_(2))V^2`

`E=(1)/(2)*50*10^(-12)*(4.2*10^(3))^2+(1)/(2)*200*10^(-12)*(4.2*10^(3))^2-(1)/(2)*(50*10^(-12)+200*10^(-12))*(4.2*10^(3))^2`

`E=0\ J`

Hence, The energy lost is zero.