CAPACITANCE
Electrostatic Potentials and Capacitance

268136 Two identical capacitorsare connected as show in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is \(\boldsymbol{q}_{0} ; \boldsymbol{k}=\) dielectric constant )

1 \(\frac{2 q_{0}}{1+\frac{1}{k}}\)
2 \(\frac{q}{1+1 / k}\)
3 \(\frac{2 q_{0}}{1+k}\)
4 \(\frac{q_{0}}{1+k}\)
Electrostatic Potentials and Capacitance

268166 In the following circuit two identical capacitors, a battery and a switch(s)are connected as shown. the switch(s) is opened and dielectric of constant \((K=3)\) are inserted in the condensers. The ratio of electrostatic energies of the system before and after filling the dielectric will be

1 \(3: 1\)
2 5
3 \(3: 5\)
4 \(5: 3\)
Electrostatic Potentials and Capacitance

272264 A parallel plate capacitor with plate area \(A\) and plate separation \(d=2 \mathrm{~m}\) has a capacitance of \(4 \mu \mathrm{~F}\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K=3\) (as shown in figure) will be:

1 \(2 \mu \mathrm{~F}\)
2 \(32 \mu \mathrm{~F}\)
3 \(6 \mu \mathrm{~F}\)
4 \(8 \mu \mathrm{~F}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Electrostatic Potentials and Capacitance

268136 Two identical capacitorsare connected as show in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is \(\boldsymbol{q}_{0} ; \boldsymbol{k}=\) dielectric constant )

1 \(\frac{2 q_{0}}{1+\frac{1}{k}}\)
2 \(\frac{q}{1+1 / k}\)
3 \(\frac{2 q_{0}}{1+k}\)
4 \(\frac{q_{0}}{1+k}\)
Electrostatic Potentials and Capacitance

268166 In the following circuit two identical capacitors, a battery and a switch(s)are connected as shown. the switch(s) is opened and dielectric of constant \((K=3)\) are inserted in the condensers. The ratio of electrostatic energies of the system before and after filling the dielectric will be

1 \(3: 1\)
2 5
3 \(3: 5\)
4 \(5: 3\)
Electrostatic Potentials and Capacitance

272264 A parallel plate capacitor with plate area \(A\) and plate separation \(d=2 \mathrm{~m}\) has a capacitance of \(4 \mu \mathrm{~F}\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K=3\) (as shown in figure) will be:

1 \(2 \mu \mathrm{~F}\)
2 \(32 \mu \mathrm{~F}\)
3 \(6 \mu \mathrm{~F}\)
4 \(8 \mu \mathrm{~F}\)
Electrostatic Potentials and Capacitance

268136 Two identical capacitorsare connected as show in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is \(\boldsymbol{q}_{0} ; \boldsymbol{k}=\) dielectric constant )

1 \(\frac{2 q_{0}}{1+\frac{1}{k}}\)
2 \(\frac{q}{1+1 / k}\)
3 \(\frac{2 q_{0}}{1+k}\)
4 \(\frac{q_{0}}{1+k}\)
Electrostatic Potentials and Capacitance

268166 In the following circuit two identical capacitors, a battery and a switch(s)are connected as shown. the switch(s) is opened and dielectric of constant \((K=3)\) are inserted in the condensers. The ratio of electrostatic energies of the system before and after filling the dielectric will be

1 \(3: 1\)
2 5
3 \(3: 5\)
4 \(5: 3\)
Electrostatic Potentials and Capacitance

272264 A parallel plate capacitor with plate area \(A\) and plate separation \(d=2 \mathrm{~m}\) has a capacitance of \(4 \mu \mathrm{~F}\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K=3\) (as shown in figure) will be:

1 \(2 \mu \mathrm{~F}\)
2 \(32 \mu \mathrm{~F}\)
3 \(6 \mu \mathrm{~F}\)
4 \(8 \mu \mathrm{~F}\)