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.

(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 .

(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 V/m.

(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 V/m.

Energy stored between the plates is .

Step 1:

(a)

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.

(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 .

(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 V/m.

(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 V/m.

Energy stored between the plates is .