Question

AN AIR CAPACITOR OF CAPITANCE C=10MICRO F IS NOW CONNECTED TO A CONSTANT VOLTAGE BATTERY OF 12V. NOW THE SPACE BETWEEN THE PLATES IS FILLED WITH A LIQUID OF A DIELECTRIC CONSTANT 5. WHAT IS THE ADDITIONAL CHARGE THAT FLOWS NOW FROM THE BATTERY TO THE CAPACITOR?

Answer 1

We know, Q = CV

So, initially, Q = 10 × 10^-3 × 12 = 0.12 C

When a dielectric is added between the plates the capacitance increases by a factor equal to the dielectric constant of the dielectric medium.

So, new charge is, Q^/ = (kC)V = k(CV) = 5 × 0.12 = 0.6 C

So, additional charge is = 0.6 – 0.12 = 0.48 C