Question

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

Answer 1

Hint:
Here we need to remember the basics of electrostatics which are used in case of capacitors. Basic formulas of electrostatic energy and relation between charge, capacitance and potential are enough to solve this question.

Step by step solution:
We will start by writing down what information is given: \[{{C}_{1}}=4\mu F=4\times {{10}^{-6}}F\]is charged using ${{V}_{1}}=200V.$

Using the formula for Electrostatic energy stored in a capacitor, $E=\dfrac{1}{2}C{{V}^{2}}.$
Initial energy of the capacitor ${{C}_{1}}$is, ${{E}_{1}}=\dfrac{1}{2}(4\times {{10}^{-6}}){{(200)}^{2}}.$

\[{{E}_{1}}=8\times {{10}^{-2}}J.\]

Now, we consider using this charged capacitor to charge another uncharged capacitor ${{C}_{1}}=2\mu F=2\times {{10}^{-6}}F.$Since, our initial capacitor is charged (That means, there is a potential difference created between the plates of the capacitor), it also produces a potential to charge the uncharged capacitor to a potential${{V}_{2}}.$

Using the conservation of charge formula, where the total charge in the circuit will be conserved, total charge in the circuit initially will be equal to the total charge in the circuit after the new capacitor is added too.

That is,${{C}_{1}}{{V}_{1}}=({{C}_{1}}+{{C}_{2}}){{V}_{2}}.$
$4\times {{10}^{-6}}\times 200=(4+2)\times {{10}^{-6}}\times {{V}_{2}}$
${{V}_{2}}=\dfrac{400}{3}V.$

Therefore, the energy of the combination of these 2 capacitors will be ${{E}_{2}}=\dfrac{1}{2}({{C}_{1}}+{{C}_{2}})V_{2}^{2}.$

${{E}_{2}}=\dfrac{1}{2}(4+2)\times {{10}^{-6}}\times {{(\dfrac{400}{3})}^{2}}$
${{E}_{2}}=5.33\times {{10}^{-2}}J.$

Therefore, the amount of energy lost by the initial capacitor${{C}_{1}}$is${{E}_{1}}-{{E}_{2}}.$
${{E}_{1}}-{{E}_{2}}=(8-5.33)\times {{10}^{-2}}J$
${{E}_{1}}-{{E}_{2}}=2.67\times {{10}^{-2}}J.$

Therefore, the amount of electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation is$2.67\times {{10}^{-2}}J.$

Note:
It is important to remember here that in the second case, when the charged capacitor is charging the uncharged capacitor there isn’t any external potential in the circuit. Hence, both the capacitors will have a common potential which was found out using the conservation of charge law.