Can you please show work! Thanks 1. A parallel plate capacitor has plates of are

Question

Can you please show work! Thanks 1. A parallel plate capacitor has plates of area 250 cm2 and separation d 2.00 cm. The capacitor is charged to a potential difference v 120 V. Then the battery is disconnected (the charge Q on the plates won't change), and a dielectric sheet (k 3.25) of the same area A but thickness l 5.00 mm is placed between the plates as shown in the figure below Determine: a. The initial capacitance of the air filled capacitor. b. The charge on each plate before the dielectric is inserted. c. The charge induced on each face of the dielectric after it is inserted. d. The electric field in the space between each plate and the dielectric. e. The electric field in the dielectric f. The potential difference between the plates after the dielectric is added g. The capacitance after the dielectric is in place

Explanation / Answer

A =250 cm^2, d = 2cm , Vo = 120 V, k =3.25, l = 5 mm

eo =8.8*10^-12 C^2/N.m^2

(a) C =Aeo/d

C = (250*10^-4*8.85*10^-12)/(0.02) =1.1*10^-11 F

C = 11 uF

(b) Qo = VoCo

Qo = 120*1.1*10^-11 = 1.32*10^-9 C

(c) Q = 1.32*10^-9C

charge is not chnaged due to dielectric

(d) Eo = Q/Aeo = (1.32*10^-9)/(250*10^-4*8.85*10^-12)

Eo = 5966 V/m

E = Q/Akeo

E = (1.32*10^-9)/(250*10^-4*3.25*8.85*10^-12)

E = 1836 V/m

Eo -E = (5966-1836) = 4130 V/m

(e) E = 1836 V/m

(f) C = keoA/(kd - l(k-1)

C = (3.25*8.85*10^-12*250*10^-4)/((3.25*0.02)-0.005(3.25-1))

C = 1.34*10^-11 F

Q =VC

V = Q/C = (1.31*10^-9)/(1.34*10^-11)

V = 97.76 V

g) C =1.34*10^-11 F

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