Figure shows two identical capacitors `C_1` and `C_2` each of `1 mu F` capacitance connected to a battery of 6 V. initially switch S is closed. After sometime S is left open and dielectric slab of dielectric constant k = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the Charge between the plates of the capacitors be affected after the slabs are inserted?

Figure shows two identical capacitors C_1 and C_2 each of 1 mu F capacitance connected to a battery of 6 V. initially switch S is closed. After sometime S is left open and dielectric slab of dielectric constant k = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the potential difference between the plates of the capacitors be affected after the slabs are inserted?

Two dielectric slabs of dielectric constant K_1 and K_2 are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. What will be the capacitance of the capacitor

Two dielectric slabs of dielectric constants K_1 and K_2 are filled in between two plates each of area A, of a parallel plate capacitor. Find net capacitance.

The potentials of the two plates of capacitors are +10 V and - 10V. The charge on one of the plates is 40 C. the capacitance of the capacitor is

Prove that if a conducting slab wholly fills the space between the plates, then the capacitance of a parallel plate capacitor is infinity.

Three capacitors C_1,C_2 and C_3 are connected to a battery of 6 V as shown in the figure find the charge on three capacitor

A parallel plate capacitor has a capacitance of 2muF . A slab of dielectric constant 5 is inserted between the plates and the capaitor is charged to 100 V and then isolated. What is the new potential difference, if the dielectric slab is removed?