Two identical capacitors are connected as shown in fignre. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. What will be the capacitance, the charge, potential difference and stored energy for each capacitor?

Two identical capacitors are connected as shown. A dielectric slab Is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. Compare the capacitance, the charge, potential difference and stored energy of each capacitor before and arter introducting of dielectric slab.

Two identical capacitors are connected as shown 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 q_(0) , k= dielectric constant). .

While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then

If the capacitor is kept connected with the battery and then the dielectric slab is inserted between the plates, then what will be the change in the charge, the capacitance the potential difference, the electric field an the stored energy?

A capacitor of capacitance C is connected to a battery of V volts. A dielectric k=(7)/(3) is introduced betwene the plates of the capacitor when battery is connected.The ratio of energies stored in the capacitor without dielectric and with dielectric..

A parallel plate capacitor is connected across a source of covstant potential difference. If a dielectric slab is introduced between the two plates, then

Two identical capacitors are connected in series with a source of constant voltage V . If Q is the charge on one of the capacitors, the capacitance of each capacitor is :

Separation between the plates of a capacitor is increased from d to 2d while the battery remains connected with the plates. How will the stored energy in the capacitor change?