Effect of Dielectric Charging and Discharging of Capacitor
Capacitance

166002 A parallel plate capacitor has area 2 m2 separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 mm, 0.6 mm,1.2 mm respectively. The capacitance is

1 5×108 Farad
2 11×108 Farad
3 2.95×108 Farad
4 10×108 Farad
Capacitance

166003 A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant K is completely inserted between the plates, then the final charge on the capacitor will be:

1 ε0AdV
2 kε0 A d V
3 ε0 AkdV
4 zero
Capacitance

166004 A capacitor of capacitance 9nF having dielectric slab of εr=2.4, dielectric strength 20 MV/m and P.D. =20 V. Calculate the area of plates.

1 2.1×104 m2
2 4.2×104 m2
3 1.4×104 m2
4 2.4×104 m2
Capacitance

166005 A capacitor of capacitance 15nF have dielectric slab of dielectric constant 2.5 , electric field strength 30MV/m and potential difference 30 volt. Calculate the area of plate.

1 6.7×104 m2
2 4.2×104 m2
3 8.0×104 m2
4 9.85×104 m2
Capacitance

166002 A parallel plate capacitor has area 2 m2 separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 mm, 0.6 mm,1.2 mm respectively. The capacitance is

1 5×108 Farad
2 11×108 Farad
3 2.95×108 Farad
4 10×108 Farad
Capacitance

166003 A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant K is completely inserted between the plates, then the final charge on the capacitor will be:

1 ε0AdV
2 kε0 A d V
3 ε0 AkdV
4 zero
Capacitance

166004 A capacitor of capacitance 9nF having dielectric slab of εr=2.4, dielectric strength 20 MV/m and P.D. =20 V. Calculate the area of plates.

1 2.1×104 m2
2 4.2×104 m2
3 1.4×104 m2
4 2.4×104 m2
Capacitance

166005 A capacitor of capacitance 15nF have dielectric slab of dielectric constant 2.5 , electric field strength 30MV/m and potential difference 30 volt. Calculate the area of plate.

1 6.7×104 m2
2 4.2×104 m2
3 8.0×104 m2
4 9.85×104 m2
Capacitance

166006 When capacitor is fully charged, find current drawn from the cell.

1 2 mA
2 1 mA
3 3 mA
4 9 mA
Capacitance

166002 A parallel plate capacitor has area 2 m2 separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 mm, 0.6 mm,1.2 mm respectively. The capacitance is

1 5×108 Farad
2 11×108 Farad
3 2.95×108 Farad
4 10×108 Farad
Capacitance

166003 A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant K is completely inserted between the plates, then the final charge on the capacitor will be:

1 ε0AdV
2 kε0 A d V
3 ε0 AkdV
4 zero
Capacitance

166004 A capacitor of capacitance 9nF having dielectric slab of εr=2.4, dielectric strength 20 MV/m and P.D. =20 V. Calculate the area of plates.

1 2.1×104 m2
2 4.2×104 m2
3 1.4×104 m2
4 2.4×104 m2
Capacitance

166005 A capacitor of capacitance 15nF have dielectric slab of dielectric constant 2.5 , electric field strength 30MV/m and potential difference 30 volt. Calculate the area of plate.

1 6.7×104 m2
2 4.2×104 m2
3 8.0×104 m2
4 9.85×104 m2
Capacitance

166006 When capacitor is fully charged, find current drawn from the cell.

1 2 mA
2 1 mA
3 3 mA
4 9 mA
Capacitance

166002 A parallel plate capacitor has area 2 m2 separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 mm, 0.6 mm,1.2 mm respectively. The capacitance is

1 5×108 Farad
2 11×108 Farad
3 2.95×108 Farad
4 10×108 Farad
Capacitance

166003 A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant K is completely inserted between the plates, then the final charge on the capacitor will be:

1 ε0AdV
2 kε0 A d V
3 ε0 AkdV
4 zero
Capacitance

166004 A capacitor of capacitance 9nF having dielectric slab of εr=2.4, dielectric strength 20 MV/m and P.D. =20 V. Calculate the area of plates.

1 2.1×104 m2
2 4.2×104 m2
3 1.4×104 m2
4 2.4×104 m2
Capacitance

166005 A capacitor of capacitance 15nF have dielectric slab of dielectric constant 2.5 , electric field strength 30MV/m and potential difference 30 volt. Calculate the area of plate.

1 6.7×104 m2
2 4.2×104 m2
3 8.0×104 m2
4 9.85×104 m2
Capacitance

166006 When capacitor is fully charged, find current drawn from the cell.

1 2 mA
2 1 mA
3 3 mA
4 9 mA
Capacitance

166002 A parallel plate capacitor has area 2 m2 separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 mm, 0.6 mm,1.2 mm respectively. The capacitance is

1 5×108 Farad
2 11×108 Farad
3 2.95×108 Farad
4 10×108 Farad
Capacitance

166003 A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant K is completely inserted between the plates, then the final charge on the capacitor will be:

1 ε0AdV
2 kε0 A d V
3 ε0 AkdV
4 zero
Capacitance

166004 A capacitor of capacitance 9nF having dielectric slab of εr=2.4, dielectric strength 20 MV/m and P.D. =20 V. Calculate the area of plates.

1 2.1×104 m2
2 4.2×104 m2
3 1.4×104 m2
4 2.4×104 m2
Capacitance

166005 A capacitor of capacitance 15nF have dielectric slab of dielectric constant 2.5 , electric field strength 30MV/m and potential difference 30 volt. Calculate the area of plate.

1 6.7×104 m2
2 4.2×104 m2
3 8.0×104 m2
4 9.85×104 m2
Capacitance

166006 When capacitor is fully charged, find current drawn from the cell.

1 2 mA
2 1 mA
3 3 mA
4 9 mA