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.
= 26.3 * 
= 2.63 *
N
