Question

A 50.0 nF air-filled parallel plate capacitor is connected in series with a 2.85 kn resistor and a battery of emf 36.0 V. (a) Determine the final charge on the capacitor. (b) Once the capacitor is fully charged and there is no current in the circuit, a sheet of plastic with a dielectric constant of 2.75 is introduced in the gap between the plates of the capacitor, completely filling the gap. If the area of each plate is 0.275 m² what is the electric field inside the plastic immediately after it is inserted (before there is any change in the charge on the plates of the capacitor)? N/C

Answer 1

Part a)

`Q = 1.8 * 10^(-6) C`

Part b)

`E = 2.69 * 10^5 N/m^2`

Step-by-step explanation:

Part a)

When capacitor is completely charged then its charge is given as

`Q = CV`

here we know that

C = 50 nF

V = 36.0 Volts

now we have

`Q = 50 * 10^(-9) * 36`

`Q = 1.8 * 10^(-6) C`

Part b)

now we know that

`C = (\epsilon_0 A)/(d)`

`50 * 10^(-9) = ((8.85 * 10^(-12))(0.275))/(d)`

`d = 4.87 * 10^(-5) m`

Now electric field inside the plastic sheet is given as

`E = (Q)/(k\epsilon_0 A)`

`E = (1.8 * 10^(-6))/(2.75 (8.85 * 10^(-12))0.275)`

`E = 2.69 * 10^5 N/m^2`