Question

The plates of a parallel-plate capacitor are 700 cm2 in area and 0.3 cm apart. The potential difference between the plates is 935 V.
(a) What is the magnitude of the electric field between the plates?
(b) the charge on each plate?
(c) the force exerted by the field on one of the plates? Now the two plates touch at the bottom forming a V-shape.
(d) Find the total electric energy stored between the plates

Answer 1

Answer: a) E = 311,666.7 V/m, b) q = 1.54×10^-10c, c) F = 4.799×10^-5 N, d) 0.0629 J

Explanation:

Area = 700cm² = (700/100)² = 7² = 49m²

Distance between plates (d) = 0.3cm = 0.3/100 = 0.003 m

V = potential difference = 935v

A)

Recall that for a capacitor that V = Ed

Where E = strength of electric field.

935 = E× 0.003

E = 935/ 0.003

E = 311,666.7 V/m

B)

C =qV

Where q = magnitude of charge on capacitor.

We need to get the value for C before we can get for q.

C =ε0×A/d

Where ε0 =permittivity of free space = 8.85×10^-12

Hence we have that

C = 8.85×10^-12 × 49/ 0.003

C = 4.32×10^-10/0.003

C = 1.44×10^-7 F

But C=qV

1.44×10^-7 = q (935)

q = 1.54×10^-10c

C)

F=Eq

Where E = strength of electric field = 311,666.7 V/m, q = 1.54×10^-10c

F = 311,666.7 × 1.54×10^-10

F = 4.799×10^-5 N

D)

Total energy stored = cv²/2

Total energy = 1.44×10^-7 × 935²/2

Total energy = 0.0629 J