Question

You've got some sheet steel and slice it up into two 16 cm (on a side) square pieces. You set them up 3.7 mm apart, then hook up a 12 V car battery, one plate to each terminal. Figure out the following list of things about your setup: (a) capacitance; (b) charge on each plate; (c) Electric field in between the plates; (d) stored energy. Unhook the battery and scoot the plates twice as far apart: what are your new answers for these four things? What if you left the battery connected when you moved the plates farther apart (same four things)?

Answer 1

a = side of square = 16 cm = 0.16 m

A = area of square = a² = 0.16 x 0.16 = 0.0256 m²

d = distance between plates = 3.7 mm = 0.0037 m

V = potential difference between the plates = 12 volts

a)

Capacitance is given as

C = _o A /d = (8.85 x 10^-12) (0.0256) / (0.0037) = 6.12 x 10^-11 F

b)

charge on each plate is given as

q = CV = (6.12 x 10^-11) (12) = 7.344 x 10^-10 C

c)

electric field between the plates is given as

E = V/d = 12 / 0.0037 = 3243.24 N/C

d)

energy stored = (0.5) C V² = (0.5) (6.12 x 10^-11) (12)² = 4.41 x 10^-9 C

E)

capacitance depends inversly on distance between plates. hence if plates are made twice apart, capacitance becomes half.

new capacitance = C/2 = 6.12 x 10^-11 /2 = 3.06 x 10^-11 F

F)

charge remains same

g)

New potential difference between plates after removing the battery and making plates twice apart is given as

C_old V_old = C_new V_new

C (12) = (C/2) V_new

V_new = 24 volts

new electric field = E_new = V_new /2d = 24 / 2d = 3243.24 N/C

d)

stored energy = (0.5) C_new V_new² = (0.5) (3.06 x 10^-11) (24)² = 8.813 x 10^-9 J