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

Answer:

Part a)

$Q = 1.8 \times 10^{-6} C$

Part b)

$E = 2.69 \times 10^5 N/m^2$

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

$Q = 50 \times 10^{-9} \times 36$

$Q = 1.8 \times 10^{-6} C$

Part b)

now we know that

$C = \frac{\epsilon_0 A}{d}$

$50 \times 10^{-9} = \frac{(8.85 \times 10^{-12})(0.275)}{d}$

$d = 4.87 \times 10^{-5} m$

Now electric field inside the plastic sheet is given as

$E = \frac{Q}{k\epsilon_0 A}$

$E = \frac{1.8 \times 10^{-6}}{2.75 (8.85 \times 10^{-12})0.275}$

$E = 2.69 \times 10^5 N/m^2$