# What happens to the charge and amount of energy stored?

In summary: This means that the charge on the plates is also halved, and therefore the energy stored by the capacitor is halved as well. So in summary, when the distance between the plates of a charged parallel plate capacitor is doubled, the capacitance is halved, the charge on the plates is halved, and the energy stored by the capacitor is also halved.

## Homework Statement

A charged parallel plate capacitor is connected to a battery. The plates of the capacitor are pulled apart so that their separation doubles. What happens to the charge and the amount of energy stored by the capacitor?

## Homework Equations

Q = cV
Ue = (1/2)(Q)^2 / c
$\Delta V = Qd/A\varepsilon_0$
$C = A\varepsilon_0 / d$

## The Attempt at a Solution

If $Q = c\Delta V$

and plugging in for c

$Q = \Delta V A \varepsilon_0 / d$

Is our original Q, now if D is separated by 2, we get

$(1/2)Q = \Delta V A \varepsilon_0 / 2d$, thus our Q has been halved.

Now if potential energy = (1/2c) (Q)^2

so doubling our distance halves our capacitance. 2d = (1/2) c

which in turn, increases our potential energy by a factor of 2.

But my book is telling me Ue is halved. Why is this??

I think I can alter the equation

Ue = (1/2c) (Q)^2

to

Ue = (1/2) c(V)^2

and then plugging in for C you get that the energy is halved.

But I just don't get how for one equation its halved, then for the other its doubled.

V can also be expressed by distance too, so the mindfuck adds on there.

Like for finding when D is doubled, finding charge,

Q = cV

well, C = Aepsilon /d , but V = Qd/ Aepsilon

So doubling the separation wouldn't do a thing here since the d's cancel out

I'm pretty confused guys. If anyone can point me in the right direction I would really appreciate it rn.

Is it maybe that I'm mistaken, and that $\Delta V$ actually doesn't change, since it depends on the batteries voltage no matter what even if the distance is being changed between the plates?

Is it maybe that I'm mistaken, and that $\Delta V$ actually doesn't change, since it depends on the batteries voltage no matter what even if the distance is being changed between the plates?
That's correct. The voltage across the capacitor remains unchanged, but since the distance between the plates is doubled, the capacitance is halved.

## 1. What happens to the charge when energy is stored?

When energy is stored, the charge remains the same. Energy and charge are two separate quantities and one does not affect the other. The charge is a property of the particle or object, while the energy is the ability to do work.

## 2. Does the amount of energy stored depend on the charge?

No, the amount of energy stored does not depend on the charge. The amount of energy stored is determined by other factors such as the type of energy storage device and the amount of charge that can be stored in it.

## 3. Can the amount of charge and energy stored change over time?

Yes, the amount of charge and energy stored can change over time. This can happen due to factors such as the flow of electric current, chemical reactions, or leakage from the storage device. However, the total amount of charge and energy in a closed system will remain constant, as per the law of conservation of charge and energy.

## 4. How can the charge and energy stored be measured?

The charge can be measured using a device called an electrometer, which measures the electric potential difference between two points. The energy stored can be measured using instruments such as a voltmeter, ammeter, or wattmeter, depending on the type of energy being stored.

## 5. What happens to the charge and energy stored when it is released?

When the energy stored is released, the charge remains the same. However, the energy is converted into a different form, such as heat, light, or mechanical motion, depending on the type of energy storage device. The charge may also flow from one point to another, resulting in an electric current.

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