A 400 pF capacitor, charged by a 100 volt d.c supply is disconnected from the supply and connected to another uncharged 400 pF capacitor. Calculate the loss of energy.

Options

A. `10^(-5)J`
B. `10^(-6)J`
C. `10^(-7)J`
D. `10^(-4)J`

Answer

B

Explanation

Given, capacitor `C=400xx10^(-12)F and V=100V`
Initially energy stored `=(1)/(2)CV^(2)=(1)/(2)[400xx10^(-12)xx(10^(2))^(2)]`
`=(1)/(2)xx4xx10^(-10)xx10^(4)=2xx10^(-6)J`
`rArr" Charge, "Q=CV=4xx10^(-10)xx100=4xx10^(-8)C`
By again connecting the charged capacitor wih an uncharged ideal capacitor.

Now, the charge Q will divide into two capacitors.
As V' and C are same for both, Q' will be same for both.
So, `Q'=(Q)/(2)=2xx10^(-8)C`
So, energy stored in both the capacitors
`=(((Q')^(2))/(2C))=((Q')^(2))/(2C)`
`=((2xx10^(-8))^(2))/(8xx10^(-10))=10^(-6)J`
`=0.5xx10^(-6)J`
So, energy lost `=2xx10^(-6)-0.5xx10^(-6)`
`=1.5xx10^(-6)J~~10^(-6)J`