Question

A parallel-plate capacitor has plates with area 0.0225 m2 separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss’s law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss’s law to calculate the electric field if the voltage source is disconnected and the Teflon is removed

Answer 1

A)

Capacitance of the capacitor is given by

C_k=Ke_oA/d =2.1*(8.85*10^-12)*(0.0225)/(1*10^-3) =4.181625*10^-10 F

Charge on each plate

Q=C_kV=(4.181625*10^-10)(12)

Q=5.018*10^-9 C

B)

From gauss' law

KEA =Q/e_o

E=Q/e_oKA =(5.018*10^-9)/(8.85*10^-12)(2.1)(0.0225)

E=12000 N/C

C)

if dielectric Teflon removed with voltage source disconnected then charge remains constant.capacitance without dielectric is

C=C_k/K =(4.181625*10^-10)/2.1 =1.99125*10^-10F

Electric filed

E=Q/e_oA =(5.018*10^-9)/(8.85*10^-12)(0.0225)

E=25200 N/C