Two capacitors of capacitance `C` and `nC` are connected in parallel. A battery of emf `V` is connected across the combination. Now the battery is removed and a dielectric constant K is inserted filling the space between the plates of capacitor of capacitance C. The final potential difference across the system is
Two capacitor C_1 with capacitance 2C and C_2 with capacitance C are connected in parallel. They are changed and then the batter is removed. If a material of dielectric constant K is inserted in C_2 . Find final potential across them.
A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,
Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential differences across the capacitors now becomes...........
Two identical parallel plate capacitors are connected in series and then joined in series will a batter of 100 V . A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitor will be, respectively.
Two identical parallel plate capacitors are connected in series to a battery of 100V . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively
A parallel-plate capacitor of plate area A and plate separation d is charged by a ideal battery of e.m.f. V and then the battery is disconnected. A slab of dielectric constant 2k is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the change in potential energy of the system in the process of inserting the slab.
Figure shows two identical capacitors C_1 and C_2 each of 2muF capacitance, connected to a battery of 5 V. Initially switch 'S' is closed. After some time 'S' is left open and dielectric slavs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?
