Remarks The answer to part (d) could also have been obtained from the electric f
ID: 1525356 • Letter: R
Question
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 electric field remains the same. The capacitance is doubled. The electric field between the plates is doubled. The electric field between the plates is halved. The charge is halved. 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
Given
area of parallel plate capacitor A = 2.30 * 10-4 m2 and
plate separation d = 1.20 *10-3 m.
a) capacitance C = epsilon not *A/d
C = (8.85*10^-12*2.3*10^-4)/(1.2*10^-3) F
C = 1.69625*10^-12 F
b) charge of the positive plate capacitor
Q = C*V = 1.69625*10^-12*3 C = 5.08875*10^-12 C
c)surface charge density sigma = Q/A = (5.08875*10^-12)/(2.30 * 10-4) C/m2
= 2.6782894736842*10^-13 C/m2
d)
magnitude of the electric field between the plates is E = sigma/epsilon not
E = (2.6782894736842*10^-13)/(8.85*10^-12) N/C
E = 0.0302631578947 N/C
E = 30.263157894 mN/C
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Given
Area A = 3.4*10^-4 m2, C = 1*10^-12 F
a) separation of the plates is d = ?
form
C = epsilon not *A/d
d = epsilon not *A/c
d = 8.85*10^-12*3.4*10^-4/1*10^-12
d = 30.09*10^-4 m
b) charge density is sigma = q/A
= 4.9*10^-12/3.4*10^-4 C/m2
= 1.4411764705882*10^-8 C/m2
c)the electric field between the plates is E = sigma/epsilon not
E = (the electric field between the plates is E = sigma/epsilon not
E = (1.4411764705882*10^-8)/(8.85*10^-12) N/C
E = 1628.45 N/C
d)
From Q = c*V
V = Q/C
V = 4.9*10^-12/1*10^-12 = 4.9 V