RC Circuits- Charge and EnergyRL vs RC circuitsWhat happens to half of the energy in a circuit with a capacitor?Capacitors as Energy Wasting?Significance of Time constants in LR and RC circuitsDifferential Equation Of Capacitor Energy In RC and RL Circuits?How does a resistor affect the voltage on a capacitor?Simple question about RC CircuitsPoynting's vector and its application to circuitsEnergy loss in capacitor for slow charge transferHeat developed in a circuit
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RC Circuits- Charge and Energy
RL vs RC circuitsWhat happens to half of the energy in a circuit with a capacitor?Capacitors as Energy Wasting?Significance of Time constants in LR and RC circuitsDifferential Equation Of Capacitor Energy In RC and RL Circuits?How does a resistor affect the voltage on a capacitor?Simple question about RC CircuitsPoynting's vector and its application to circuitsEnergy loss in capacitor for slow charge transferHeat developed in a circuit
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It takes longer for the charge in an RC circuit to fall down to half its initial value than for the stored energy. Why?
electric-circuits electrical-resistance capacitance
edited 7 hours ago
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It takes longer for the charge in an RC circuit to fall down to half its initial value than for the stored energy. Why?
electric-circuits electrical-resistance capacitance
edited 7 hours ago
Qmechanic♦
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asked 8 hours ago
AaradhanbAaradhanb
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$begingroup$
It takes longer for the charge in an RC circuit to fall down to half its initial value than for the stored energy. Why?
electric-circuits electrical-resistance capacitance
edited 7 hours ago
Qmechanic♦
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asked 8 hours ago
AaradhanbAaradhanb
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It takes longer for the charge in an RC circuit to fall down to half its initial value than for the stored energy. Why?
electric-circuits electrical-resistance capacitance
electric-circuits electrical-resistance capacitance
edited 7 hours ago
Qmechanic♦
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asked 8 hours ago
AaradhanbAaradhanb
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edited 7 hours ago
Qmechanic♦
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edited 7 hours ago
Qmechanic♦
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edited 7 hours ago
Qmechanic♦
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edited 7 hours ago
Qmechanic♦
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asked 8 hours ago
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asked 8 hours ago
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3 Answers
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$begingroup$
The energy stored in the electric field of a capacitor is given by
$$E=fracCV^22$$
The relationship between voltage, charge and capacitance is
$$C=fracQV$$
Substituting for V is the first equation
$$E=fracQ^22C$$
Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.
Hope this helps.
answered 8 hours ago
Bob DBob D
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$begingroup$
Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
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The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.
But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on
- How much charge you move, and
- The force you need to apply to each charge to get it to move to the other plate
Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.
Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.
answered 7 hours ago
Aaron StevensAaron Stevens
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The energy stored in the electric field of a capacitor is given by
$$E=fracCV^22$$
The relationship between voltage, charge and capacitance is
$$C=fracQV$$
Substituting for V is the first equation
$$E=fracQ^22C$$
Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.
Hope this helps.
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
The energy stored in the electric field of a capacitor is given by
$$E=fracCV^22$$
The relationship between voltage, charge and capacitance is
$$C=fracQV$$
Substituting for V is the first equation
$$E=fracQ^22C$$
Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.
Hope this helps.
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
The energy stored in the electric field of a capacitor is given by
$$E=fracCV^22$$
The relationship between voltage, charge and capacitance is
$$C=fracQV$$
Substituting for V is the first equation
$$E=fracQ^22C$$
Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.
Hope this helps.
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
$endgroup$
The energy stored in the electric field of a capacitor is given by
$$E=fracCV^22$$
The relationship between voltage, charge and capacitance is
$$C=fracQV$$
Substituting for V is the first equation
$$E=fracQ^22C$$
Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.
Hope this helps.
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
answered 8 hours ago
Bob DBob D
10.1k3 gold badges9 silver badges34 bronze badges
10.1k3 gold badges9 silver badges34 bronze badges
add a comment |
add a comment |
$begingroup$
Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
New contributor
Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
New contributor
Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
New contributor
Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
New contributor
Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
New contributor
Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
answered 8 hours ago
Leiba GoldsteinLeiba Goldstein
266 bronze badges
266 bronze badges
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Leiba Goldstein is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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add a comment |
add a comment |
$begingroup$
The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.
But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on
- How much charge you move, and
- The force you need to apply to each charge to get it to move to the other plate
Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.
Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
$endgroup$
add a comment |
$begingroup$
The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.
But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on
- How much charge you move, and
- The force you need to apply to each charge to get it to move to the other plate
Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.
Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
$endgroup$
add a comment |
$begingroup$
The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.
But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on
- How much charge you move, and
- The force you need to apply to each charge to get it to move to the other plate
Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.
Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
$endgroup$
The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.
But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on
- How much charge you move, and
- The force you need to apply to each charge to get it to move to the other plate
Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.
Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
answered 7 hours ago
Aaron StevensAaron Stevens
17.7k4 gold badges29 silver badges65 bronze badges
17.7k4 gold badges29 silver badges65 bronze badges
add a comment |
add a comment |
Aaradhanb is a new contributor. Be nice, and check out our Code of Conduct.
Aaradhanb is a new contributor. Be nice, and check out our Code of Conduct.
Aaradhanb is a new contributor. Be nice, and check out our Code of Conduct.
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