Effect of Dielectric Charging and Discharging of Capacitor
Capacitance

165968 The plates of a parallel plate capacitor are charged upto $200 \mathrm{~V}$. A dielectric slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

1 1
2 4
3 5
4 6
Capacitance

165969 A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $50 \mathrm{~V}$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 \mathrm{~V}$. The capacitance of second capacitor is

1 $10 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $30 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165970 The function of a dielectric in a capacitor is

1 to reduce the effective potential on plates
2 to increase the effective potential on plates
3 to reduce the capacitor of plate
4 to decrease the capacitance
Capacitance

165971 An air filled parallel plate capacitor charged to potential ' $V_{1}$ ' is connected to uncharged parallel plate capacitor having dielectric constant ' $K$ '. The common potential of both is ' $V_{2}$ '. The value of ' $K$ ' will be

1 $\frac{V_{1}-V_{2}}{V_{1}+V_{2}}$
2 $\frac{V_{1}-V_{2}}{V_{2}}$
3 $\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
4 $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{1}-\mathrm{V}_{2}}$
Capacitance

165972 The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in SI unit is $\varepsilon_{0}=$ $8.85 \times 10^{-12}$ SI unit

1 15
2 $8.845 \times 10^{-11}$
3 $0.4425 \times 10^{-10}$
4 5
Capacitance

165968 The plates of a parallel plate capacitor are charged upto $200 \mathrm{~V}$. A dielectric slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

1 1
2 4
3 5
4 6
Capacitance

165969 A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $50 \mathrm{~V}$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 \mathrm{~V}$. The capacitance of second capacitor is

1 $10 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $30 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165970 The function of a dielectric in a capacitor is

1 to reduce the effective potential on plates
2 to increase the effective potential on plates
3 to reduce the capacitor of plate
4 to decrease the capacitance
Capacitance

165971 An air filled parallel plate capacitor charged to potential ' $V_{1}$ ' is connected to uncharged parallel plate capacitor having dielectric constant ' $K$ '. The common potential of both is ' $V_{2}$ '. The value of ' $K$ ' will be

1 $\frac{V_{1}-V_{2}}{V_{1}+V_{2}}$
2 $\frac{V_{1}-V_{2}}{V_{2}}$
3 $\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
4 $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{1}-\mathrm{V}_{2}}$
Capacitance

165972 The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in SI unit is $\varepsilon_{0}=$ $8.85 \times 10^{-12}$ SI unit

1 15
2 $8.845 \times 10^{-11}$
3 $0.4425 \times 10^{-10}$
4 5
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Capacitance

165968 The plates of a parallel plate capacitor are charged upto $200 \mathrm{~V}$. A dielectric slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

1 1
2 4
3 5
4 6
Capacitance

165969 A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $50 \mathrm{~V}$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 \mathrm{~V}$. The capacitance of second capacitor is

1 $10 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $30 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165970 The function of a dielectric in a capacitor is

1 to reduce the effective potential on plates
2 to increase the effective potential on plates
3 to reduce the capacitor of plate
4 to decrease the capacitance
Capacitance

165971 An air filled parallel plate capacitor charged to potential ' $V_{1}$ ' is connected to uncharged parallel plate capacitor having dielectric constant ' $K$ '. The common potential of both is ' $V_{2}$ '. The value of ' $K$ ' will be

1 $\frac{V_{1}-V_{2}}{V_{1}+V_{2}}$
2 $\frac{V_{1}-V_{2}}{V_{2}}$
3 $\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
4 $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{1}-\mathrm{V}_{2}}$
Capacitance

165972 The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in SI unit is $\varepsilon_{0}=$ $8.85 \times 10^{-12}$ SI unit

1 15
2 $8.845 \times 10^{-11}$
3 $0.4425 \times 10^{-10}$
4 5
Capacitance

165968 The plates of a parallel plate capacitor are charged upto $200 \mathrm{~V}$. A dielectric slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

1 1
2 4
3 5
4 6
Capacitance

165969 A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $50 \mathrm{~V}$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 \mathrm{~V}$. The capacitance of second capacitor is

1 $10 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $30 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165970 The function of a dielectric in a capacitor is

1 to reduce the effective potential on plates
2 to increase the effective potential on plates
3 to reduce the capacitor of plate
4 to decrease the capacitance
Capacitance

165971 An air filled parallel plate capacitor charged to potential ' $V_{1}$ ' is connected to uncharged parallel plate capacitor having dielectric constant ' $K$ '. The common potential of both is ' $V_{2}$ '. The value of ' $K$ ' will be

1 $\frac{V_{1}-V_{2}}{V_{1}+V_{2}}$
2 $\frac{V_{1}-V_{2}}{V_{2}}$
3 $\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
4 $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{1}-\mathrm{V}_{2}}$
Capacitance

165972 The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in SI unit is $\varepsilon_{0}=$ $8.85 \times 10^{-12}$ SI unit

1 15
2 $8.845 \times 10^{-11}$
3 $0.4425 \times 10^{-10}$
4 5
Capacitance

165968 The plates of a parallel plate capacitor are charged upto $200 \mathrm{~V}$. A dielectric slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

1 1
2 4
3 5
4 6
Capacitance

165969 A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $50 \mathrm{~V}$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20 \mathrm{~V}$. The capacitance of second capacitor is

1 $10 \mu \mathrm{F}$
2 $20 \mu \mathrm{F}$
3 $30 \mu \mathrm{F}$
4 $15 \mu \mathrm{F}$
Capacitance

165970 The function of a dielectric in a capacitor is

1 to reduce the effective potential on plates
2 to increase the effective potential on plates
3 to reduce the capacitor of plate
4 to decrease the capacitance
Capacitance

165971 An air filled parallel plate capacitor charged to potential ' $V_{1}$ ' is connected to uncharged parallel plate capacitor having dielectric constant ' $K$ '. The common potential of both is ' $V_{2}$ '. The value of ' $K$ ' will be

1 $\frac{V_{1}-V_{2}}{V_{1}+V_{2}}$
2 $\frac{V_{1}-V_{2}}{V_{2}}$
3 $\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}}$
4 $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{1}-\mathrm{V}_{2}}$
Capacitance

165972 The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in SI unit is $\varepsilon_{0}=$ $8.85 \times 10^{-12}$ SI unit

1 15
2 $8.845 \times 10^{-11}$
3 $0.4425 \times 10^{-10}$
4 5