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.) By what fraction does the capacitance change? C_new= (a) (b) By what fraction does the amount of charge change? Q_new= (c) By what fraction does the voltage difference between the two plates change? AV_new=| C_old; ] Q_old (When the battery is disconnected, the charges have no place to go to and will stay where they were). ]av_old (When the battery is disconnected, voltage difference may no longer be kept to be the same value any more.). (d) By what fraction does the energy stored in the capacitor change? U_new= | Does it make sense? Let's check, considering the amount of charge and the area of the plate, (e) by what fraction does the amount of charge density change? a_new=| (f) Consider how E is related to the charge density, by what fraction does the E field change? E new= U_old ]e old E_old. Considering how E, d and voltage difference are related, does it make sense now, by what fraction does the Voltage difference change? Please explain why the energy stored in the capacitor changes? If it increased, where did the energy come from? If it decreased, where could the energy go? (The force between the two plates are attracting or repelling?. When the plate separation decreases, does the electric force do positive work or negative work?)
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.) By what fraction does the capacitance change? C_new= (a) (b) By what fraction does the amount of charge change? Q_new= (c) By what fraction does the voltage difference between the two plates change? AV_new=| C_old; ] Q_old (When the battery is disconnected, the charges have no place to go to and will stay where they were). ]av_old (When the battery is disconnected, voltage difference may no longer be kept to be the same value any more.). (d) By what fraction does the energy stored in the capacitor change? U_new= | Does it make sense? Let's check, considering the amount of charge and the area of the plate, (e) by what fraction does the amount of charge density change? a_new=| (f) Consider how E is related to the charge density, by what fraction does the E field change? E new= U_old ]e old E_old. Considering how E, d and voltage difference are related, does it make sense now, by what fraction does the Voltage difference change? Please explain why the energy stored in the capacitor changes? If it increased, where did the energy come from? If it decreased, where could the energy go? (The force between the two plates are attracting or repelling?. When the plate separation decreases, does the electric force do positive work or negative work?)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
Transcribed Image Text: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 to go to and will stay where they were).
(c) By what fraction does the voltage difference between the two plates change? AV_new=
|AV_old
(When the battery is disconnected,voltage difference may no longer be kept to be the same value any more.).
(d) By what fraction does the energy stored in the capacitor change? U_new=
U_old
Does it make sense? Let's check, considering the amount of charge and the area of the plate,
(e) by what fraction does the amount of charge density change? o_new=
0_old
(f) Consider how E is related to the charge density, by what fraction does the E field change? E_new=
E_old.
Considering how E, d and voltage difference are related, does it make sense now, by what fraction does the Voltage difference change?
Please explain why the energy stored in the capacitor changes? If it increased, where did the energy come from? If it decreased, where could the energy go? (The force between the two plates are attracting or repelling?. When the plate
separation decreases, does the electric force do positive work or negative work?)
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