EXAMPLE 16.6 A Parallel-Plate Capacitor SOLUTION (A) Find the capacitance. Subst

Question

EXAMPLE 16.6 A Parallel-Plate Capacitor

SOLUTION

(A) Find the capacitance.

Substitute values into the definition of capacitance:

(B) Find the charge on the positive plate after the capacitor is connected to a 3.00-V battery.

Substitute into the capacitance charge equation to find the charge:

(C) Calculate the charge density on the positive plate.

Charge density is charge divided by area:

(D) Calculate the magnitude of the electric field between the plates.

Apply V = Ed.

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REMARKS The answer to part (d) could also have been obtained from the electric field derived for a parallel plate capacitor E = /0.

QUESTION How do the following change if the distance between the plates is doubled? (Select all that apply.)

The charge is halved.The capacitance is doubled.The electric field between the plates is doubled.The electric field between the plates is halved.The electric field remains the same.The charge is doubled.The capacitance is halved.

PRACTICE IT

(a) Find its capacitance.
F

(b) How much charge is on the positive plate if the capacitor is connected to a 3.00 V battery?
C

(c) Calculate the charge density on the positive plate, assuming the density is uniform.
C/m2

(d) Calculate the magnitude of the electric field between the plates.
V/m

EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!

Explanation / Answer

here,

area of plate, A = 2.30 * 10^-4 m^2
plate seperation, d = 1.20 * 10^-3 m

Part a:
capcacitance is parallel plate capacitot is given as :
C = eo*A/d
C = (8.85*10^-12 * 2.30*10^-4)/(1.20*10^-3)
C = 1.696*10^-12 F

Part b :
Voltage of battery, V = 3 V
Charge, Q = C * V
Q = 1.696*10^-12 * 3
Q = 5.088*10^-12 C

Part C:
Electric Field on plate is given as , E = Sigma/ Epsilon
also Electric field is E = V/d

Sigma = (V/d)*eo
sigma = (3 / (1.20 * 10^-3)) * 8.85*10^-12
sigma = 2.213*10^-8 C/m^3

Part D:
Electric Field Between the plates,

E = V/d
E = (3 / (1.20 * 10^-3))
E = 2500 V/m

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