Question

A parallel plate capacitor with separation, d, is charged up to a surface charge density σ0. a) The plates are now disconnected from the charging device. The separation between the plates is now increased from d to D (D > d). How much work per unit area is done by the agent causing the increase in separation? b) When the plates are at a separation, d, and charge density σ0, they are attached to a battery that maintains the voltage ﬁxed at the value corresponding to (d, σ0). Once again the surfaces are separated from d to D, but this time with the voltage ﬁxed. How much work per unit area is performed by the agent causing the increase in separation? Discuss this result in the context of (a) above.

Answer #1

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= Q_old (When the battery is disconnected, the
charges have no place...

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A dielectric-filled parallel-plate capacitor has plate area A =
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8.85×10−12 C2/N⋅m2 .
a. The dielectric plate is now slowly pulled out of the
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Throughout the problem, use ϵ0 = 8.85×10−12
C2/N⋅m2 .
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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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