What's the energy stored in a parallel plate capacitor?

[NTesla]

Homework Statement

I'm studying about parallel plate capacitor. I have a question about the energy stored in it. The book says the energy stored is given by ${1/2}CV{^2}$.

However, when I'm trying to reason in a different way, I'm facing problem. Here's how I'm doing it: $F = QE$.
$Q = CV$.
$V = Ed$,
and Work done = $F.d$ = $QEd$ =$CV. E. d$ =$CV. {V/d} . d$ =$CV{^2}$.

So, my question is where did the other half of $CV{^2}$ go?

Relevant Equations

F = QE, W =Fd.

I've mentioned the problem that i'm facing in the question that I've posted. Kindly go through it.

 

[haruspex]

$F = QE$.
$Q = CV$.
$V = Ed$,
and Work done = $F.d$ = $QEd$ =$CV. E. d$ =$CV. {V/d} . d$ =$CV{^2}$.

You need to give reasons for applying a formula.
What physical process are you considering and what do all those variables mean in the context of that process?

 

[NTesla]

F = QE, is the force experienced by one plate of capacitor by the another plate. Q is the magnitude of charge on the plate which is experiencing the force. E is the electric field between the plates of the capacitor.

Q = CV is the usual expression for the charge, capacitance and the voltage between the plates.

V = Ed is the equation between voltage difference between the plates and the Electric field, d is the distance between the plates.

Work done = F. d is the work done by the external agent. The reasoning is that initially the plates are almost touching each other. Then the external agent does work on the capacitor while pulling the plates apart to a certain distance d. That work done is stored in the capacitor as energy.

Here's the relevant part of the book which I'm reading:

 

[haruspex]

F = QE, is the force experienced by one plate of capacitor by the another plate. Q is the magnitude of charge on the plate which is experiencing the force. E is the electric field between the plates of the capacitor.

Q = CV is the usual expression for the charge, capacitance and the voltage between the plates.

V = Ed is the equation between voltage difference between the plates and the Electric field, d is the distance between the plates.

Work done = F. d is the work done by the external agent. The reasoning is that initially the plates are almost touching each other. Then the external agent does work on the capacitor while pulling the plates apart to a certain distance d. That work done is stored in the capacitor as energy.

Here's the relevant part of the book which I'm reading: View attachment 351214 View attachment 351215

In post #1 you used E=V/d, but d is not constant during the process you are considering.

 

[NTesla]

Yes d is not constant. But same can also be said about Capacitance and voltage. The way it has been detailed in the book, capacitance and voltage has been taken as constant. The way the energy stored has been calculated in the book, is it wrong ?

 

[Gavran]

Yes d is not constant. But same can also be said about Capacitance and voltage. The way it has been detailed in the book, capacitance and voltage has been taken as constant. The way the energy stored has been calculated in the book, is it wrong ?

The book is correct.
What is the work done by the person during the displacement from $x$ to $x+dx$ where $dx$ is an infinitesimally small amount of displacement?

 

[Steve4Physics]

Yes d is not constant. But same can also be said about Capacitance and voltage. The way it has been detailed in the book, capacitance and voltage has been taken as constant. The way the energy stored has been calculated in the book, is it wrong ?

What the book says is correct. It specifically states that ‘C is the capacitance of the capacitor in the final position’. It follows that V and U are the voltage and energy in the final position. C, V and U are not used as intermediate (variable) values.

The force on a capacitor plate is not $F = QE$. It is $F = \frac 12 QE$. You need to show this as part of your overall derivation. This will then give you the required result.

However, your derivation needs improvements as already noted by @haruspex.

 

[haruspex]

Yes d is not constant. But same can also be said about Capacitance and voltage. The way it has been detailed in the book, capacitance and voltage has been taken as constant. The way the energy stored has been calculated in the book, is it wrong ?

There is the field generated by one plate, $E_1$, and the field between the plates, $E_2=2E_1$.
Your $E$ refers to one of these at one point and the other at another.

Edit: @Steve4Physics spotted it already.

 

[haruspex]

The book is correct.
What is the work done by the person during the displacement from $x$ to $x+dx$ where $dx$ is an infinitesimally small amount of displacement?

Yes, I went down that path, but it's not the error. See posts #7 and #8.

 

[kuruman]

Saying

F = QE, is the force experienced by one plate of capacitor by the another plate.

is inconsistent with the underlying model that treats the plate under consideration as if it were a point charge $Q$ in an external field $E$. If that were the case, then the other plate should also be treated as a point charge and write, consistently but incorrectly, the attractive Coulomb force as $$F=\frac{1}{4\pi\epsilon_0}\frac{Q^2}{d^2}.$$

 

[NTesla]

There is the field generated by one plate, $E_1$, and the field between the plates, $E_2=2E_1$.
Your $E$ refers to one of these at one point and the other at another.

Edit: @Steve4Physics spotted it already.

Thank you @haruspex, @Steve4Physics. Your comment were helpful. Much appreciated.

 

[anuttarasammyak]

8.2 of https://www.feynmanlectures.caltech.edu/II_08.html might help and assure your understanding.

 

[Gavran]

Yes, I went down that path, but it's not the error. See posts #7 and #8.

The approach in the post #6 is correct.
$dW=F(x)dx$ where $F(x)$ is the force applied by the first plate on the second plate and $F(x)=E(x)Q$ where $E(x)$ is the electric field generated by the first plate. $Q$ is a charge on the second plate.

 

[haruspex]

The approach in the post #6 is correct.
$dW=F(x)dx$ where $F(x)$ is the force applied by the first plate on the second plate and $F(x)=E(x)Q$ where $E(x)$ is the electric field generated by the first plate. $Q$ is a charge on the second plate.

Yes, post #6 is a perfectly valid way to solve the textbook problem, but that is not what the OP is asking for. The OP wants to know what is wrong with the method in post #1.
Posts #7 and #8 answer that.

 

[Gavran]

Yes, post #6 is a perfectly valid way to solve the textbook problem, but that is not what the OP is asking for. The OP wants to know what is wrong with the method in post #1.
Posts #7 and #8 answer that.

okay