if a 5microfarad capacitor is charged by a 20volts battery and it is then connected across an uncharged 20microfarad capacitor , find the potential energy
Find the PE of what?
5*10^-6 = Q/20
Q = 10^-4
That charge is distributed on the new capacitance of C = (5+20)10^-6 = 25 *10^-6
V = Q/C
V = 10^-4/(25*10^-6) = 10^2/25 = 4 Volts
Energy = (1/2) Q V = .5 *10^-4 * 4
= 2*10^-4 Joules
To find the potential energy in this scenario, we need to calculate the total charge stored in the capacitors and then use the formula for potential energy in a capacitor.
First, let's find the charge on the 5 microfarad capacitor.
We know that the capacitance (C) of the first capacitor is 5 microfarads (µF), and the voltage (V) across it is 20 volts (V). The formula to calculate the charge (Q) stored in a capacitor is: Q = C * V.
Q1 = 5 µF * 20 V = 100 µC
Now, when the charged 5 microfarad capacitor is connected across an uncharged 20 microfarad capacitor, the charge will distribute itself between the two capacitors.
Since the capacitors are connected in parallel, the total charge remains the same. Therefore, the charge on the 20 microfarad capacitor will also be 100 µC.
Now, let's calculate the potential energy.
The formula for potential energy (PE) stored in a capacitor is: PE = (1/2) * C * V^2.
For the 5 microfarad capacitor:
PE1 = (1/2) * 5 µF * (20 V)^2 = 1000 µJ
For the 20 microfarad capacitor:
PE2 = (1/2) * 20 µF * (20 V)^2 = 8000 µJ
Finally, to find the total potential energy, we can add the potential energies of both capacitors:
Total PE = PE1 + PE2 = 1000 µJ + 8000 µJ = 9000 µJ
Therefore, the potential energy in this system is 9000 microjoules (µJ).