Effect of Dielectric Charging and Discharging of Capacitor
Capacitance

166042 If initial capacitance of capacitor is $C$, then final capacity of capacitor will be

1 $\mathrm{KC}$
2 $(\mathrm{K}+1) \mathrm{C}$
3 $(\mathrm{K}+1) \mathrm{C} / 2$
4 $(\mathrm{K}-1) \mathrm{C}$
Capacitance

166043 When two identical capacitors are charged individually to different potentials and then connected in parallel, after disconnecting from the source

1 net charge $=$ sum of initial charges
2 net potential difference $\neq$ sum of individual initial potential difference
3 net energy stored $<$ sum of individual initial energy
4 All of the above
Capacitance

166044 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \mathrm{~cm}$ is $2 \mu \mathrm{F}$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8 . The final capacity of the capacitor is

1 $11.2 \mu \mathrm{F}$
2 $15.6 \mu \mathrm{F}$
3 $19.2 \mu \mathrm{F}$
4 $22.4 \mu \mathrm{F}$
Capacitance

166046 A parallel plate capacitor has a capacity $80 \times$ $10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

1 $45.6 \times 10^{-3} \mathrm{C}$
2 $25.3 \times 10^{-3} \mathrm{C}$
3 $120 \times 10^{-3} \mathrm{C}$
4 $125 \times 10^{-3} \mathrm{C}$
Capacitance

166042 If initial capacitance of capacitor is $C$, then final capacity of capacitor will be

1 $\mathrm{KC}$
2 $(\mathrm{K}+1) \mathrm{C}$
3 $(\mathrm{K}+1) \mathrm{C} / 2$
4 $(\mathrm{K}-1) \mathrm{C}$
Capacitance

166043 When two identical capacitors are charged individually to different potentials and then connected in parallel, after disconnecting from the source

1 net charge $=$ sum of initial charges
2 net potential difference $\neq$ sum of individual initial potential difference
3 net energy stored $<$ sum of individual initial energy
4 All of the above
Capacitance

166044 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \mathrm{~cm}$ is $2 \mu \mathrm{F}$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8 . The final capacity of the capacitor is

1 $11.2 \mu \mathrm{F}$
2 $15.6 \mu \mathrm{F}$
3 $19.2 \mu \mathrm{F}$
4 $22.4 \mu \mathrm{F}$
Capacitance

166046 A parallel plate capacitor has a capacity $80 \times$ $10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

1 $45.6 \times 10^{-3} \mathrm{C}$
2 $25.3 \times 10^{-3} \mathrm{C}$
3 $120 \times 10^{-3} \mathrm{C}$
4 $125 \times 10^{-3} \mathrm{C}$
Capacitance

166042 If initial capacitance of capacitor is $C$, then final capacity of capacitor will be

1 $\mathrm{KC}$
2 $(\mathrm{K}+1) \mathrm{C}$
3 $(\mathrm{K}+1) \mathrm{C} / 2$
4 $(\mathrm{K}-1) \mathrm{C}$
Capacitance

166043 When two identical capacitors are charged individually to different potentials and then connected in parallel, after disconnecting from the source

1 net charge $=$ sum of initial charges
2 net potential difference $\neq$ sum of individual initial potential difference
3 net energy stored $<$ sum of individual initial energy
4 All of the above
Capacitance

166044 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \mathrm{~cm}$ is $2 \mu \mathrm{F}$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8 . The final capacity of the capacitor is

1 $11.2 \mu \mathrm{F}$
2 $15.6 \mu \mathrm{F}$
3 $19.2 \mu \mathrm{F}$
4 $22.4 \mu \mathrm{F}$
Capacitance

166046 A parallel plate capacitor has a capacity $80 \times$ $10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

1 $45.6 \times 10^{-3} \mathrm{C}$
2 $25.3 \times 10^{-3} \mathrm{C}$
3 $120 \times 10^{-3} \mathrm{C}$
4 $125 \times 10^{-3} \mathrm{C}$
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Capacitance

166042 If initial capacitance of capacitor is $C$, then final capacity of capacitor will be

1 $\mathrm{KC}$
2 $(\mathrm{K}+1) \mathrm{C}$
3 $(\mathrm{K}+1) \mathrm{C} / 2$
4 $(\mathrm{K}-1) \mathrm{C}$
Capacitance

166043 When two identical capacitors are charged individually to different potentials and then connected in parallel, after disconnecting from the source

1 net charge $=$ sum of initial charges
2 net potential difference $\neq$ sum of individual initial potential difference
3 net energy stored $<$ sum of individual initial energy
4 All of the above
Capacitance

166044 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \mathrm{~cm}$ is $2 \mu \mathrm{F}$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8 . The final capacity of the capacitor is

1 $11.2 \mu \mathrm{F}$
2 $15.6 \mu \mathrm{F}$
3 $19.2 \mu \mathrm{F}$
4 $22.4 \mu \mathrm{F}$
Capacitance

166046 A parallel plate capacitor has a capacity $80 \times$ $10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

1 $45.6 \times 10^{-3} \mathrm{C}$
2 $25.3 \times 10^{-3} \mathrm{C}$
3 $120 \times 10^{-3} \mathrm{C}$
4 $125 \times 10^{-3} \mathrm{C}$