Question

A parallel-plate capacitor is charged by connection to a battery. If the battery is disconnected and the separation between the plates is increased, what will happen to the charge on the capacitor and the voltage across it?

A) Both remain fixed.

B) Both Increase.

C) Both decrease.

D) The charge increases and the voltage decreases.

E) The charge remains fixed and the voltage increases.

Answer #1

a
parallel plate capacitor is fully charged by a battery, and then
the battery is disconnected. if a dielectric material is then
inserted between the capacitor plates, which of the following
quantities would remain constant?
A) electric field between the plates
B) potential difference across the plates
C) charge on each plate
D) energy stored in capacitor
E) charge and energy would both remain constant

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A parallel-plate capacitor is charged and then disconnected from
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Then, the plate separation is decreased by a factor of 4, d_new=1/4
d_old,
please type in a number into the box to determine how everything
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(If something doesn't change, please type in 1 to answer X_new= 1
X_old.)
(a) By what fraction does the capacitance change?
C_new= C_old;
(b) By what fraction does the amount of charge change?
Q_new= Q_old (When the battery is disconnected, the
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The new potential difference between the plates
The Field between the plates after increasing the separation
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The voltage across the capacitor is doubled.
The charge on the plates is doubled.
The charge on the plates decreases by a factor of 2.
The electric field is doubled.

A parallel plate capacitor with plate separation d is connected
to a battery. The capacitor is fully charged to Q Coulombs and a
voltage of V. (C is the capacitance and U is the stored energy.)
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
dielectric with κ...

A parallel-plate capacitor is charged and then disconnected from
a battery.
Then, the plate separation is decreased by a factor of 4, d_new=1/4
d_old,
please type in a number into the box to determine how everything
changes.
(If something doesn't change, please type in 1 to answer X_new= 1
X_old.)
(a) By what fraction does the capacitance change? C_new=
_________________ C_old;
(b) By what fraction does the amount of charge change? Q_new=
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A parallel plate capacitor with plate separation d is connected
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