A 40uf parallel plate capacitor is charged to 12v. The capacitor is removed from the battery and the separation distance increased from 2mm to 3mm.

What is the total charge on the capacitor?

How much energy was stored in the capacitor when the distance was 2mm?

By how much is the energy changed when there plate separation is increased to 3mm.

To find the total charge on the capacitor, we can use the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the voltage.

Given that the capacitance is 40µF (or 40x10^-6 F) and the voltage is 12V, we can substitute these values into the formula:

Q = (40x10^-6 F) * 12V
= 0.00048 C

Therefore, the total charge on the capacitor is 0.00048 Coulombs.

To calculate the energy stored in the capacitor when the distance was 2mm, we can use the formula:

E = (1/2) * C * V^2

where E is the energy, C is the capacitance, and V is the voltage.

Substituting the values into the formula:

E = (1/2) * (40x10^-6 F) * (12V)^2
= 0.00288 J

Therefore, the energy stored in the capacitor when the distance was 2mm is 0.00288 Joules.

To find the change in energy when the plate separation is increased to 3mm, we can subtract the energy at 3mm from the energy at 2mm.

Let's calculate the energy at 3mm first using the same formula:

E = (1/2) * C * V^2
E = (1/2) * (40x10^-6 F) * (12V)^2
= 0.00216 J

Now, let's find the change in energy:

Change in energy = Energy at 3mm - Energy at 2mm
= 0.00216 J - 0.00288 J
= -0.00072 J

Therefore, the change in energy when the plate separation is increased to 3mm is -0.00072 Joules.