How Does Changing the Air Gap Affect the Energy Stored in a Capacitor?

[hellogirl88]

A capacitor is completely charged with 650 nC by a voltage source that had 375 V.

The initial air gap of the capacitor above was 7 mm. What is the stored energy if the air gap is now 11 mm?

 

[LowlyPion]

Has the voltage source been disconnected before increasing the distance?

 

[nlightner]

The energy stored in a capacitor can be calculated using the equation E = 1/2 * C * V^2, where E is the stored energy, C is the capacitance, and V is the voltage.

In this scenario, the capacitor is initially charged with 650 nC by a voltage source of 375 V. The initial air gap is 7 mm, but it is now 11 mm. To calculate the stored energy, we first need to determine the capacitance of the capacitor. This can be done using the equation C = Q/V, where Q is the charge and V is the voltage. In this case, the capacitance would be 650 nC/375 V = 1.73 nF.

Now, using the equation E = 1/2 * C * V^2, we can calculate the stored energy with the new air gap of 11 mm. Plugging in the values, we get:

E = 1/2 * (1.73 nF) * (375 V)^2 = 153.4 nJ

Therefore, the stored energy in the capacitor is 153.4 nJ when the air gap is 11 mm. This is an increase from the initial stored energy of the capacitor when the air gap was 7 mm. This is because the capacitance decreases as the air gap increases, resulting in a higher stored energy.