A parallel-plate capacitor has the capacitance of C 1 = 20 ?F when the distance

Question

A parallel-plate capacitor has the capacitance of C1 = 20 ?F when the distance of the two plates is d = 5 cm. A battery with the emf = 18 V is used to charge the capacitor.

1. How much charge can be put on the parallel-plate capacitor? (in the unit of ?C)

Now, consider the following two cases separately:
Case 1: if the emf stays connected, but the distance of the parallel plates is changed to d1 = 0.5 cm.
2. What is the new capacitance C1 and how much is the charge Q1in the capacitor?

Case 2: After the initial charging, the emf is removed. Therefore, the amount of charge you calculated from the question #1 remains in the capacitor. Now, again, the distance of the parallel plates is changed to d1 = 0.5 cm.
3. What is the potential difference V1 across the capacitor?

Explanation / Answer

We know that

The capacitance of the capacitor is given by

               C =eoA/d

Then area between the plates is given by

               A =C*d/eo =20*10-6*5*10-2/8.85*10-12 =11.29*104m2

Now charge on the parallel palte capacitor is Q =CV =(20*10-6)(18) =360uC

If the distance between the plates is reduced to d =0.5cm then d =0.5*10-2m

Now the new capacitance is given by

            C =eoA/d =8.85*10-12*11.29*104m2/0.5*10-2m=199.833uF =200uF

now the charge is Q =CV =(200uF)(18V) =3600uC = 3.60mC

After the initial charging, the emf is removed. Therefore, the amount of charge you calculated from the question #1 remains in the capacitor. Now, again, the distance of the parallel plates is changed to d1 = 0.5 cm.

Then the potential is given by

                           Q =CV then V =Q/C =3600uC/200uF =18V

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