Question

A parallel plate capacitor has plates of area A =0.055
m^{2} separated by distance d =2.20 ✕ 10^{−5} m and
is attached to a battery with potential difference ΔV = 19 V.(The
permittivity of free space is ε_{o}= 8.85 ✕
10^{−12} C^{2}/N·m^{2}).

(a) Calculate the capacitance (in F) if the space between the plates is filled with air.

(b) What is the electric field between the plates of the capacitor? In N/C

**Suppose that this capacitor is replaced with another
with capacitance C = 2.20 ✕ 10 ^{−5} F.**

(c) What is the energy stored in the new capacitor? in Joules

(d) The charging battery is now disconnected and a ceramic slab with dielectric constant κ = 2.30 is slipped between the plates of the new capacitor. What is the energy stored in the capacitor now?

Answer #1

A parallel plate capacitor has plates of area A =0.075
m2 separated by distance d =2.30 ✕ 10−5 m and
is attached to a battery with potential difference ΔV = 14 V.(The
permittivity of free space is εo= 8.85 ✕
10−12 C2/N·m2).
(a)Calculate the capacitance (in F) if the space between the
plates is filled with air.
(b) What is the electric field between the plates of the
capacitor?
Suppose that this capacitor is replaced with another
with capacitance C...

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is εo= 8.85 ✕ 10−12 C2/N·m2). (a) Calculate the capacitance (in F)
if the space between the plates is filled with air. F (b) What is
the electric field between the plates of the capacitor? Suppose
that this capacitor is replaced with another with...

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calculate the capacitance in F if the space between the plates
is filled with air
what is the electric field between the plates of the
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suppose the capacitor is replaced with another capacitance C=
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A parallel-plate capacitor has plates of area 0.16 m2 and a
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is then
disconnected. A dielectric slab of thickness 0.4 cm and dielectric
constant K=3 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?
c) What is the free charge q before the slab is...

A parallel plate capacitor is made from two plates 7 cm^2 in
area, with a plate separation of 3 mm. The capacitor is fully
charged across a 30 V battery, and then disconnected.
How much charge is on the capacitor?
Now material with a dielectric constant of 30 is inserted to
fill the space between the plates, (with the battery disconnected),
What is the charge on the capacitor?
What is the energy stored in the capacitor?
What the field between...

A parallel plate capacitor, in which the space between the
plates is filled with a dielectric material with dielectric
constant κ = 4.7, has a capacitance of C=1,018 μF and it is
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Once it is fully charged, it is disconnected from the battery and
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How much change occurs in the energy of the capacitor;...

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charges ±Q , which are initially separated by a distance d . The
capacitor is not connected to a battery. Then, the plates are moved
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dielectric with dielectric constant κ=2 that fills the entire space
in the gap is inserted. What happens to the stored energy U and the
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A parallel plate capacitor, in which the space between the
plates is filled with a dielectric material with dielectric
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A parallel plate capacitor with plate separation d is connected
to a battery. The capacitor is fully charged to Q Coulombs and a
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Answer the following questions regarding the capacitor charged by a
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After being disconnected from the battery, inserting a
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After being disconnected from the battery, inserting a
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An air-filled parallel-plate capacitor with capacitance
C0 stores charge Q on plates separated by
distance d. The potential difference across the plates is
ΔV0 and the energy stored is PEC,0. If the
capacitor is disconnected from its voltage source and the space
between the plates is then filled with a dielectric of constant k =
2.00, evaluate the ratios
(a) Cnew/C0,
(b) ΔVnew/ ΔV0, and
(c) PEC, new/PEC, 0.

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