Question

1) Find the electric field inside a 10 microFarad parallel plate capacitor when connected to a 6V battery if the gap between the capacitor plates is filled with air. Repeat your calculation if the gap is filled with paper with a dielectric constant of 4?

2) A capacitor is formed between two metal plates, both 10 x 10cm, separated by 1mm. A) What is its capacitance? B) When connected to a 10 V battery how much charge flows onto the plates? C) What is the net electric field between the plates? D) What is the total force on each plate?

Answer 1

2)

A)

A = area of each plate = 10 x 10 = 100 cm2 = 100 x 10-4 m2

d = separation between the plates = 1 mm = 1 x 10-3 m

Capacitance is given as

C = A/d = (8.85 x 10-12) (100 x 10-4)/(1 x 10-3) =

C = 8.85 x 10-11 F

B)

V = potential difference across the plates = 10 volts

charge is given as

Q = CV

Q = ( 8.85 x 10-11) (10)

Q = 8.85 x 10-10 C

C)

Electric field between the plates is given as

E = V/d = 10/(1 x 10-3) = 1 x 104 N/C

D)

Total force on each plate is given as

F = Q E

F = ( 8.85 x 10-10) (1 x 104)

F = 8.85 x 10-6 N