Parallel plate capacitor's plate distance is doubled, does the energy double?

In summary, in a parallel-plate capacitor, when the separation between the plates is doubled, the electric field remains unchanged because it is not affected by the distance between the plates. However, the potential difference and total energy both double because they are directly proportional to the distance between the plates. This may seem counterintuitive, but it can be explained by the equations relating capacitance, charge, and voltage.
  • #1
DerekZ10
4
0

Homework Statement



A parallel-plate capacitor is charged by being connected to a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric field change? The potential difference? The total energy? Explain your reasoning.

Homework Equations



σ=Q/A (Surface Charge Density)

E=σ/ε0 (Electric Field Between Plates)

V=Ed (Potential Difference)

U=(1/2)QV (Potential Energy)

The Attempt at a Solution



Q1=Q2

A1=A2

d1=x
d2=2x

Question 1: Electric Field Change?

E=Q/(Aε0)

No, d is not related.

Question 2: Potential Difference Change?

V=Ed

Yes, the voltage will double.

Question 3: Total Energy change?

U=(1/2)QV

Yes, it will double.


4. Hold up! That doesn't make sense!

So I have the answers, but I need someone to explain this to me because this looks to be breaking thermodynamics. The answers given is inline with what cramster and numerous other sources have said. I think the new capacitor will allow it to store twice the charge, but the problem makes it seem like the energy is just created. Or am I suppose to be assuming that by moving the plates farther apart, the capacitor is converting the physical motion energy into electrical energy?

My logic is telling me the final total energy should be the same as the initial. And that the electric field should have been halved. But I don't have an equation available for electric field intensity.
 
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  • #2
DerekZ10 said:

Homework Statement



A parallel-plate capacitor is charged by being connected to a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric field change? The potential difference? The total energy? Explain your reasoning.

Homework Equations



σ=Q/A (Surface Charge Density)

E=σ/ε0 (Electric Field Between Plates)

V=Ed (Potential Difference)

U=(1/2)QV (Potential Energy)

The Attempt at a Solution



Q1=Q2

A1=A2

d1=x
d2=2x

Question 1: Electric Field Change?

E=Q/(Aε0)

No, d is not related.

Question 2: Potential Difference Change?

V=Ed

Yes, the voltage will double.

Question 3: Total Energy change?

U=(1/2)QV

Yes, it will double.


4. Hold up! That doesn't make sense!

So I have the answers, but I need someone to explain this to me because this looks to be breaking thermodynamics. The answers given is inline with what cramster and numerous other sources have said. I think the new capacitor will allow it to store twice the charge, but the problem makes it seem like the energy is just created. Or am I suppose to be assuming that by moving the plates farther apart, the capacitor is converting the physical motion energy into electrical energy?

My logic is telling me the final total energy should be the same as the initial. And that the electric field should have been halved. But I don't have an equation available for electric field intensity.

Welcome to the PF.

It looks like you are missing two key equations. The first relates the capacitance to the palte area A and separation distance d. Are you familiar with that equation?

The second equation relates the charge Q to the capacitance C and the voltage between the plates V. Are you familiar with that equation?

Does using those two equations make this easier for you?
 

FAQ: Parallel plate capacitor's plate distance is doubled, does the energy double?

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy. It consists of two conducting plates separated by a distance, with a dielectric material in between.

2. How does changing the plate distance affect the energy stored in a parallel plate capacitor?

Changing the plate distance in a parallel plate capacitor directly affects the capacitance, which is a measure of the amount of stored energy. Increasing the distance between the plates decreases the capacitance, and thus decreases the stored energy. Conversely, decreasing the distance between the plates increases the capacitance and the stored energy.

3. If the plate distance is doubled, does the energy double?

No, doubling the plate distance does not double the energy stored in a parallel plate capacitor. As discussed earlier, the energy stored is directly proportional to the capacitance, which is inversely proportional to the plate distance. Therefore, doubling the plate distance will result in a halving of the capacitance and a decrease in the stored energy by a factor of 2.

4. What happens to the electric field when the plate distance is doubled?

When the plate distance is doubled, the electric field between the plates decreases. This is because the electric field is inversely proportional to the plate distance. As the distance increases, the electric field weakens, and vice versa.

5. Does doubling the plate distance affect the voltage in a parallel plate capacitor?

Yes, doubling the plate distance also affects the voltage in a parallel plate capacitor. The voltage between the plates is directly proportional to the electric field and the plate distance. Therefore, doubling the distance will result in a halving of the voltage and vice versa.

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