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Question

A parallel-plate capacitor with plate area A = 2.0 m2 and plate separation d = 3.0 mm is connected to a 35-V battery (Fig. 17–51a). (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17–51b). What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor? 

 

Answer 1

Step 1:

(a)

Area of the each plate is 2.0 m².

Plate separation is 3.0 mm = 3.0*10^-3 m.

Voltage applied between the plates is 35 V.

(i)

Find the capacitance.

Capacitance of the ideal parallel plate is , where A is the area of each plate, d  is the plate separation and

is the permittivity of free space = .

Capacitance of the parallel plate is 5.9 nF.

Step 2:

(ii)

Find the charge on the capacitance.

Charge on the capacitance is , where C  is the capacitance and V  is the voltage applied.

Charge on the capacitor is .

Step 3:

(iii)

Find the electric field.

Electric field between parallel plate is , where V is the potential difference between the plates and d is the plate separation.

Electric field between parallel plate is 11.67 kV/m.

Step 4:

(iv)

Find the energy stored between the plates.

Energy stored between the plates is , where V  is the applied voltage and C is the capacitance.

Energy stored between the plates is .

Solution:

Capacitance of the parallel plate is 5.9 nF.

Charge on the capacitor is .

Electric field between parallel plate is 11.67 kV/m.

Energy stored between the plates is .

Answer 2

Step 1:

(b)

Now a plastic slab is placed between the parallel plates.

Area of the each plate is 2.0 m².

Plate separation is 3.0 mm = 3.0*10^-3 m.

Voltage applied between the plates is 35 V.

Dielectric constant of the plastic slab is 3.2.

(i)

Find the capacitance.

Capacitance of the parallel plate is , where A is the area of each plate, d is the plate separation,

k is the dielectric constant of the medium and is the permittivity of free space = .

Capacitance of the parallel plate is 18.88 nF.

Step 2:

(ii)

Find the charge on the capacitance.

Charge on the capacitance is , where C is the capacitance and V is the voltage applied.

Charge on the capacitor is .

Step 3:

(iii)

Find the electric field.

Electric field between parallel plate is , where V is the potential difference between the plates and d is the plate separation.

Electric field between parallel plate is 11.67 kV/m.

Step 4:

(iv)

Find the energy stored between the plates.

Energy stored between the plates is , where V is the applied voltage and C is the capacitance.

Energy stored between the plates is .

Solution:

Capacitance of the parallel plate is 18.88 nF.

Charge on the capacitor is .

Electric field between parallel plate is 11.67 kV/m.

Energy stored between the plates is .