Question

A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to the potential difference between the plates?

Answer 1

Answer:

The potential difference reduced by half of its value 0.5V

Explanation:

Terms to Note

Potential difference

Potential difference is the difference in the potential energy of two points along a conductor carry a charge.

Capacitance

Capacitance is the measure of energy that is stored in a capacitor which is a circuit component that stores energy

Step by Step Calculation

The capacitance, Charge and potential relationship can be expressed in equation 1

Q = CV .....................1

where Q is the charge

         C is the Capacitance and

        V is the potential difference

but C = εoA/d ....................2

Substitution the value of C in equation 2 into equation 1 we have

Q = εoA/d x V

making the potential difference the subject formula we have

V = Qd/εoA ...........................3

since the separation was decreased by d/2 we have a new value for d and is substituted into equation 3  and V is given a new value as $V_{1\\}$

$V_{1\\}$ = $\frac{Qd}{2e_{o}A }$

To obtain the final potential difference we compute the ratio of V and $V_{1\\}$

V:$V_{1\\}$ = Qd/εoA : Qd/2εoA

V:$V_{1\\}$ = 1:2

$V_{1\\}$ = 0.5 V

So the new potential difference is 0.5 V  which halves of the first potential difference.

Answer 2

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2