2. A parallel-plate capacitor has a charge of 5.OuC when charged by a potential
difference of 1.25V.
a. Find its capacitance.
b. How much electrical potential energy is stored when this capacitor is
connected to a 1.5V battery?

Given a capacitor that is charge by a battery

So when the battery was disconnect, the charge remains on the capacitor

One plate will have positive charge

And the other will have a negative charge but of same magnitude

Now, the plates are pulled apart, so the distance between them was extended

a. The capacitance of the capacitor?

The capacitance of a capacitor is given as

C = εo•A/d

Where C is the capacitance

εo is permissivity, a constant

A is are of plate

d is the distance apart

From the formula we notice that

Capacitance is inversely proportional to distance, so this implies that as the distance is increase, the capacitance is reduce.

Capacitance Decreases

b. Potential difference?

The potential difference is given as

q = CV

V = q/C

Or V=Ed

Where E is electric field

So the potential difference is inversely proportional to the capacitance, so we know that the capacitance is reducing, then voltage will increase.

Or, using the second relations

V=Ed, voltage is directly proportional to distance, So, if the distance apart is increasing then, the voltage is increasing

Voltage is increasing

c. Electric field between plate.

The electric field is given as

E = kQ/r²

So the electric field is inversely proportional to the distance square, so as the distance is an increase, the electric field is decreased.

Electric field decreases

d. Electric potential energy?

U = ½CV², then, q = CV

Then, U = ½qV

The electric potential energy is directly proportional to the voltage and since the voltage is increasing due to increase in distance, then the electric potential energy increases.

reserve is the answer