Question

A parallel plate capacitor filled with mica having εr = 5 is connected to a 10 V battery. The area of each parallel plate is 6 cm2 and separation distance is 6 mm.
(a) Find the capacitance and stored charge.
(b) After the capacitor is fully charged, the battery is disconnected and the dielectric is removed carefully. Calculate the new values of capacitance, stored energy and charge.

Answer 1

Answer:

There are two parts of this question. First is related to the find of capacitance and stored charge. Second is related to the new value of capacitance, energy and charge. Therefore, we need two solutions for this particular question that are given in the explanation portion of the question.

Explanation:

Part a:

Given data:

εr = 5

Voltage =V = 10V

$Area=A=6cm^{2} =0.0006m^{2}$

Distance = d = 6mm = 0.006 m

Now as we know that

C=(A εr εo)/d

Here εo = 8.85 x 10-12 F/m

Putting the values in above equation we will get.

C = (0.0006*5*8.85 x 10-12)/0.006

C = (2.65x10-14)/0.006

C = 4.425x10-12 F

OR

C = 4.425 pF

The value of stored charge is given by the formula

Q = CV

Q = (4.425 pF)*10

Q = 44.4 nC

Part b:

As the battery is already removed and after the removal of dielectric the value of capacitance will be.

Co = C/εr

Co = 4.425 pF/5

Co = 0.855x10-12

OR

Co = 0.855 pF

The value of new stored charge will be the same as in the previous case and that is.

Q = 44.4 nC

The value of stored energy is given by the formula.

$U_o=\frac{Q^2}{2C_o}$

Putting the values

Uo = (44.4 nC*44.4 nC)/(2*0.855 pF)

Uo = 1.97x10-15/1.71x10-12

Uo = 1.15 mJ

Conclusion:

Capacitance = 4.425 pF

Charge stored = 44.4 nC

New value of capacitance = 0.855 pF

Energy stored = 1.15 mJ