2.27 A 4 \mu F capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 \mu Fcapacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answers (1)

Here,

The charge on the capacitance Initially

Q=CV=4*10^{-6}*200=8*10^{-4}C

Total electrostatic energy initially

E_{initial}=\frac{1}{2}CV^2=\frac{1}{2}4*10^{-6}*(200)^2=8*10^{-2}J

Now, when it is disconnected and connected to another capacitor

Total New capacitance = C_{new}=4+2=6\mu F

Now, by conserving the charge on the capacitor:

V_{new}C_{new}=C_{initial}V_{initial}

V_{new}6\mu F=4\mu F *200

V_{new}=\frac{400}{3}V

Now,

New Electrostatic energy :

E_{new}=\frac{1}{2}C_{new}V_{new}^2=\frac{1}{2}*6*10^{-6}*\left ( \frac{400}{3} \right )^2=5.33*10^{-2}J

Therefore,

Lost in electrostatic energy 

E=E_{initial}-E_{new}=0.08-0.0533=0.0267J

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