Question

The figure shows two identical capacitors C1andC2 each of 2μF capacitance, connected to a battery of 5 V. Initially switch 'S' is closed. After some time 'S' is left open and dielectric slabs 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?

Answer 1

Answer:

the figure is not given so I will not be able to tell the exact answer but...

Explanation:

since the capacitors are connected in series, this capacitance will be (1 /C1+1/C2) the while raised to power minus 1.

now since a dielectric is inserted between them, their capacitance will be k times the initial value.

put the formula Q is equal to CV and put new value of C and get the answer.

hope it helps.

Answer 2

Answer:

hope it is helpful to you