Problem #1
A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 100 cm2, and a mica dielectric (ϵr = 5.4) completely filling the space between the plates. At 50 V potential difference, calculate (a) the electric field magnitude E in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.Answer;
Known:
capacitance is C = 100 pF = 100 x 10─12 F
plate area is A = 100 cm2 = 100 x 10─4 m2
dielectric, ϵr = 5.4
Potential difference, V = 50 V
(a) the electric field magnitude E in the mica
The distance between the plates, d, is
d = ϵr ϵ0A/C = (5.4)(8.85 x 10─12 C2/Nm2)(100 x 10─4 m2)/(100 x 10─12 F) = 4.779 x 10─3 m
then,
E = V/d = 50 V/0.004779 m = 10462 V/m = 10.46 kV/m
(b) the magnitude of the free charge on the plates is
q = CV = (100 x 10─12 F)(50 V) = 5.0 nC
(c) the magnitude of the induced surface charge on the mica.
q' = q ─ (q/ϵr) = 5nC (1 ─ 1/5.4) = 4.1 nC
Problem #2
A parallel-plate capacitor has plates of area 0.12 m2 and a separation of 1.2 cm. A battery charges the plates to a potential difference of 120 V and is then disconnected. A dielectric slab of thickness 4.0 mm and dielectric constant 4.8 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is
the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?
Answer;
Known:
Plates of area is A = 0.12 m2
Separation is d = 1.2 cm = 1.2 x 10─2 m
Potential difference is V = 120 V
Dielectric constant, ϵr = 4.8
Dielectric slab of thickness is b = 4.0 mm = 4.0 x 10─3 m
(a) the capacitance before the slab is inserted given by
C0 = ϵ0A/d = (8.85 x 10─12 C2/Nm2)(0.12 m2)/(1.2 x 10─2 m) = 8.9 x 10─11 F = 89 pF
(b) the capacitance with the slab in place is
C = ϵrϵ0A/[ϵr(d ─ b) + b]
C = (4.8)(8.85 x 10─12 C2/Nm2)(0.12 m2)/[4.8(0.012 m ─ 0.004 m) + 0.004 m] = 1.2 x 10─10 F = 120 pF
and
(c) the free charge q before the slab is inserted is
q0 = C0V = (8.9 x 10─11 F)(120 V) = 1.1 x 10─8 C = 11 nC
(d) the free charge q after the slab is inserted is
q = q0 = 11 nC
baterai had been disconnected, meaning no new q is added.
(e) the magnitude of the electric field in the space between the plates and dielectric
E0 = q0/ϵ0A = (1.1 x 10─8 C)/(8.85 x 10─12 C2/Nm2 x 0.12 m2) = 10 kV/m
(f) the magnitude of the electric field in the dielectric itself?
E = E0/ϵr = (10 kV/m)/4.8 = 2.16 kV/m
(g) the potential difference across the plates with the slab in place is
V = E0(d ─ b) + Eb = (10 kV/m)(0.012 m ─ 0.004 m) + (2.16 kV/m)(0.004 m) = 88.64 V
(h) external work is involved in inserting the slab is
W = ΔU = q2/2[1/C ─ 1/C0] = (1.1 x 10─8 C)2/2[1/(1.2 x 10─10 F) ─ 1/(8.9 x 10─10 F)] = ─176 nJ
Problem #3
Two parallel plates of area 100 cm2 are given charges of equal magnitudes 8.9 x 10─7 C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 x 106 V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
Answer;
Known:
Charges, q = 8.9 x 10─7 C
Plates of area is A = 100 cm2 = 100 x 10─4 m2
Electric field, E = 1.4 x 106 V/m
(a) the dielectric constant of the material isϵr = q0/Eϵ0A
= (8.9 x 10─7 C)/[(1.4 x 106 V/m)(8.85 x 10─12 C2/Nm2)(100 x 10─4 m2)] =
ϵr = 7.2
(b) the magnitude of the charge induced on each dielectric surface is
q ─ q’ = q0/ϵr
where: q = induced charge and q’ = charge on plates
q’ = q(1 ─ 1/ϵr) = (8.9 x 10─7 C)(1 ─ 1/7.2) = 7.7 x 10─7 C = 0.77 μC
Problem #4
The space between two concentric conducting spherical shells of radius b = 1.70 cm and a = 1.20 cm is filled with a substance of dielectric constant ϵr = 23.5. A potential difference V = 73.0 V is applied across the inner and outer shells. Determine (a) the capacitance of the device, (b) the free charge q on the inner shell, and (c) the charge q’ induced along the surface of the inner shell.
Answer;
Known:
spherical shells of radius b = 1.70 cm = 1.70 x 10─2 m and a = 1.20 cm = 1.20 x 10─2 m
dielectric constant ϵr = 23.5
Potential difference V = 73.0 V
(a) the capacitance of the device is
C0 = 4πϵ0ab/(a ─ b)
= 4π(8.85 x 10─12 C2/Nm2)(1.70 x 10─2 m)(1.20 x 10─2 m)/[1.70 x 10─2 m ─ 1.20 x 10─2 m]
C0 = 4.537 x 10─12 F
then,
C = ϵrC0 = 23.5 x 4.537 x 10─12 F
C = 1.066 x 10─10 F = 1.07 nF
(b) the free charge q on the inner shell is
q = CV = (1.066 x 10─10 F)(73.0 V) = 7.79 nC
(c) the charge q’ induced along the surface of the inner shell
q’ = q(1 ─ 1/ϵr) = (7.79 nC)(1 ─ 1/23.5) = 7.7 x 10─7 C = 7.46 nC
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