Question

A homemade capacitor is assembled by placing two 10 in pie pans 8 cm apart and connecting them to the opposite terminals of a 9 V battery.

(a) Estimate the capacitance.
  pF
(b) Estimate the charge on each plate.
pC
(c) Estimate the electric field halfway between the plates.
V/m
(d) Estimate the work done by the battery to charge the plates.
J
(e) Which of the above values change if a dielectric is inserted? (Select all that apply.)

electric fieldchargework done by the batterycapacitance

Answer 1

given that ::

voltage of battery, V = 9 V

distance between two 10 in pie pans, d = 8 cm = 0.08 m

in a parallel plate capacitor, capacitance is given as ::

              C = (E A) / d                                                  ( eq. 1 )

where , C = capacitance of the capacitor
E = permitivity of insulator
A = area of each plate
d = distance between two pie pans

for air, E = E_r x E_o = 1.00054 x 8.85419 x 10^-12 F/m

A = r² = x 9 in² x (2.54 cm/in)² x (1 m / 100 cm)²

A = 1.82415 x 10^-2 m²

inserting the values in eq. 1,

C = (8.85897 x 10^-12 ) (1.82415 x 10^-2 m² ) / 0.08

C = 16.16 x 10^-14 / 0.08 = 202 x 10^-14 F

C = 2 x 10^-12 F = 2 pF

b) using, capacitance C = Q / V                              

or Q = C V                                                    ( eq. 2 )

where, Q = charge on each plate

V = terminal voltage

inserting values in eq.2,

Q = (2 x 10^-12 F) (9 V) = 18 x 10^-12 = 18 pC

c) Electric field is given as ::

E = V / d                                           { eq. 3 }

assume constant e field.so, halfway between plates would be 1/2 x V/d

E = (1/2) x 9 V / 0.08 m = 56.25 V/m

d) work done by the battery to charge the plates is given as ::

dW = (Q / C) x dQ                                  { eq. 4 }

integral (dW) = integral 0 to Q of [ (Q / C). dQ],

W = (1/2) Q² / C                                { eq. 5 }

but C = Q / V so, Q² = C² / V²

W = (1/2) C V²                     ( eq. 6 )

where, W = work done

C = capacitance of the capacitor

V = voltage

inserting the values in eq. 6,

W = 1/2 x (2 x 10^-12 ) ( 9 V )² = 9 x 10^-12 J

(e) capacitance values change if a dielectric is inserted because, if capacitance is changed, assuming voltage is

constant, then charge will change as well..                      Q = CV

no change in an electric field, since E = Vd, and we just assumed voltage to be constant.