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Concerning Work Required to Charge a Capacitor

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Ackbach

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Jan 26, 2012
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I'm investigating the derivation of the equation $W=CV^{2}/2$ for the amount of work it takes to charge a capacitor of capacitance $C$ to the point that the voltage across it is $V$. In Griffiths' Introduction to Electrodynamics, 4th Ed., this is Equation 2.55 on page 107. In his derivation, he says that to move a chunk of charge $dq$ against the electric field (necessary to charge a capacitor), you would have to do a chunk of work $dW=V \, dq$. But I'm not sure I buy this. He references Eq. 2.38, which is essentially that $V=W/q$, for the potential difference between two points being equal to the work per unit charge. Rearranging yields $W=qV$. But now, mathematically, if I take the differential of the LHS, I must do so to the RHS: $dW=V\, dq+q\, dV$, via the product rule. $V=q/C$, so it's not constant with respect to $q$. Why can Griffiths (and lots of other physics textbooks doing this derivation) ignore the $dV$ term? With $dV=dq/C$, I get that
$$dW=\frac{q}{C}\, dq+ \frac{q}{C} \, dq= \frac{2q \, dq}{C},$$
in which case $W=CV^{2}$, twice what the book gets.

What's wrong with this picture?
 

dwsmith

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Feb 1, 2012
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I'm investigating the derivation of the equation $W=CV^{2}/2$ for the amount of work it takes to charge a capacitor of capacitance $C$ to the point that the voltage across it is $V$. In Griffiths' Introduction to Electrodynamics, 4th Ed., this is Equation 2.55 on page 107. In his derivation, he says that to move a chunk of charge $dq$ against the electric field (necessary to charge a capacitor), you would have to do a chunk of work $dW=V \, dq$. But I'm not sure I buy this. He references Eq. 2.38, which is essentially that $V=W/q$, for the potential difference between two points being equal to the work per unit charge. Rearranging yields $W=qV$. But now, mathematically, if I take the differential of the LHS, I must do so to the RHS: $dW=V\, dq+q\, dV$, via the product rule. $V=q/C$, so it's not constant with respect to $q$. Why can Griffiths (and lots of other physics textbooks doing this derivation) ignore the $dV$ term? With $dV=dq/C$, I get that
$$dW=\frac{q}{C}\, dq+ \frac{q}{C} \, dq= \frac{2q \, dq}{C},$$
in which case $W=CV^{2}$, twice what the book gets.

What's wrong with this picture?
Have you looked at the wikipedia page?

https://en.wikipedia.org/wiki/Capacitance
 
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Ackbach

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Jan 26, 2012
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Klaas van Aarsen

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The formula $V=W/q$ is not generally true.
It only holds if the charge $q$ is small enough so it has no significant effect on the electric field.
In other words, $q$ has to be infinitely small, or really $dq$.
I do not have Griffiths', but does he perchance mention this?

To elaborate, a possible derivation is:
$$W= F s = (q E) s = q (E s) = q V$$
This derivation holds (only) if $q$ is infinitely small and if $F$ can assumed to be constant.
 
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Ackbach

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Jan 26, 2012
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The formula $V=W/q$ is not generally true.
It only holds if the charge $q$ is small enough so it has no significant effect on the electric field.
In other words, $q$ has to be infinitely small, or really $dq$.
I do not have Griffiths', but does he perchance mention this?

To elaborate, a possible derivation is:
$$W= F s = (q E) s = q (E s) = q V$$
This derivation holds (only) if $q$ is infinitely small and if $F$ can assumed to be constant.
Right. Griffiths derives $W=qV$ in the context of one particle of charge $q$ in the presence of an electric field $\mathbf{E}$ that is not, itself, altered by the charge $q$. The field $\mathbf{E}$ is not assumed constant. I will type up Griffiths' derivation here:


Suppose you have a stationary configuration of source charges, and you want to move a test charge $Q$ from point $\mathbf{a}$ to points $\mathbf{b}$. Question: How much work will you have to do? At any point along the path, the electric force on $Q$ is $\mathbf{F}=Q\mathbf{E}$; the force you must exert, in opposition to this electrical force, is $-Q\mathbf{E}$. (If the sign bothers you, think about lifting a brick; gravity exerts a force $mg$ downward, but you exert a force $mg$ upward. Of course, you could apply an even greater force - then the brick would accelerate, and part of your effort would be "wasted" generating kinetic energy. What we're interested in here is the minimum force you must exert to do the job.) The work you do is therefore
$$W= \int_{ \mathbf{a}}^{ \mathbf{b}} \mathbf{F} \cdot d\mathbf{l}
=-Q \int_{ \mathbf{a}}^{ \mathbf{b}} \mathbf{E} \cdot d\mathbf{l}
=Q[V( \mathbf{b})-V( \mathbf{a})].$$
Notice that the answer is independent of the path you take from $\mathbf{a}$ to $\mathbf{b}$; in mechanics, then, we would call the electrostatic force "conservative." Dividing through by $Q$, we have
$$V(\mathbf{b})-V( \mathbf{a})= \frac{W}{Q}. \qquad \qquad \qquad (2.38)$$
In other words, the potential difference between points $\mathbf{a}$ and $\mathbf{b}$ is equal to the work per unit charge required to carry a particle from $\mathbf{a}$ to $\mathbf{b}$.

Then, in deriving $W= \displaystyle \frac{CV^{2}}{2}$, Griffiths writes as follows:


To "charge up" a capacitor, you have to remove electrons from the positive plate and carry them to the negative plate. In doing so, you fight against the electric field, which is pulling them back toward the positive conductor and pushing them away from the negative one. How much work does it take, then, to charge the capacitor up to a final amount $Q$? Suppose that at some intermediate stage in the process the charge on the positive plate is $q$, so that the potential difference is $q/C$. According to Eq. 2.38, the work you must do to transport the next piece of charge, $dq$, is
$$dW= \left( \frac{q}{C} \right) dq.$$
The total work necessary, then, to go from $q=0$ to $q=Q$, is
$$W= \int_{0}^{Q} \left( \frac{q}{C} \right) dq= \frac{1}{2} \frac{Q^{2}}{C},$$
or, since $Q=CV,$
$$W= \frac{1}{2} CV^{2},$$
where $V$ is the final potential of the capacitor.

It's this second derivation that I am questioning.
 

Klaas van Aarsen

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It's this second derivation that I am questioning.
I'm not sure if I understand.
As Griffith writes, the formula W=QV is only applicable to a "test charge", which should be interpreted as an infinitely small charge.
So the formula from the first derivation is really dW=dQ V.
This is the formula he uses in the second derivation.

Furthermore, he also uses the formula Q/V=C being constant, which is a property of a capacitor (where Q is the total charge on a plate).
 
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Ackbach

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I'm not sure if I understand.
As Griffith writes, the formula W=QV is only applicable to a "test charge", which should be interpreted as an infinitely small charge.
So the formula from the first derivation is really dW=dQ V.
This is exactly my question. $V$ changes with $Q$, does it not? In which case, if I have $W=QV$, why is $dW = V dQ$, and not $W=V dQ+Q dV$? The second employs the product rule for differentials, whereas, it seems to me, the first does not.
 
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Ackbach

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Jan 26, 2012
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Interestingly, here is an alternate derivation that "violates" no mathematics of which I am aware:
\begin{align*}
P&=VI \\
\frac{dW}{dt}&=V \left( C \frac{dV}{dt} \right) \\
\frac{dW}{dt}&= \frac{d}{dt} \left( \frac{CV^{2}}{2} \right) \\
W&=\frac{CV^{2}}{2}.
\end{align*}
The constant of integration can be taken to be zero.

So I am now convinced that my objection is incorrect. However, I want to know, on the basis of the Griffiths derivation, why it's right and my objection is wrong.
 

Klaas van Aarsen

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if I have $W=QV$
But you do not.

why is $dW = V dQ$
This is what you have.
The first form is misleading, since it suggests it is true for an arbitrary charge Q, which it is not.

It seems I am missing something in your argument, since I feel I am repeating myself...

Edit: Note that $P=VI$ is true for arbitrary quantities and not just for infinitesimal ones.
 

Klaas van Aarsen

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Let me try a parallel.

Suppose we integrate the area $A$ under a function $f$.
The area increases by $dA=f(x)dx$, so yes, the area increases, but that does not mean that $A=f(x)x$ and therefore $dA=f'(x)xdx + f(x)dx$.
 

Ackbach

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$$dW=V \, dq$$
This is what you have.
The first form is misleading, since it suggests it is true for an arbitrary charge Q, which it is not.
Ok, I get that the charge has to be small enough so that it does not materially affect the field. With that understanding, is $W=VQ$ still wrong?

It seems I am missing something in your argument, since I feel I am repeating myself...
Let me ask this question: how would you derive $dW=V \, dq$ from first principles, similar to the first Griffiths derivation I quoted above? Would you simply start with an infinitesimal chunk of charge $dq$ and move on from there?

Let me try a parallel.

Suppose we integrate the area $A$ under a function $f$.
The area increases by $dA=f(x)dx$, so yes, the area increases, but that does not mean that $A=f(x)x$ and therefore $dA=f'(x)xdx + f(x)dx$.
I would agree, but from the perspective of the electrodynamics question before us, this seems to be assuming what I'm setting about to prove; namely, that $dW=V \, dq$.
 

Klaas van Aarsen

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Ok, I get that the charge has to be small enough so that it does not materially affect the field. With that understanding, is $W=VQ$ still wrong?
Mathematically, it only does not affect the field if it is infinitesimally small.
In practice, it is treated as such if its effect is negligible, meaning less than ~2%.
This is what is called a "test charge".



Let me ask this question: how would you derive $dW=V \, dq$ from first principles, similar to the first Griffiths derivation I quoted above? Would you simply start with an infinitesimal chunk of charge $dq$ and move on from there?
Yes.

$$W= \int_Q \int_{ \mathbf{a}}^{ \mathbf{b}} \mathbf{dF} \cdot \mathbf{dl}
=-\int_Q \int_{ \mathbf{a}}^{ \mathbf{b}} dq\ \mathbf{E} \cdot \mathbf{dl}$$


I would agree, but from the perspective of the electrodynamics question before us, this seems to be assuming what I'm setting about to prove; namely, that $dW=V \, dq$.
Consider it like this: if we move right by a small enough amount, we can consider f(x) to be constant, or rather that the value of f(x) is not significantly influenced by the change in x coordinate. Therefore the area increases by f(x)dx.
If we move right by a small amount h, we won't in good consciousness be able to say that the area increases by f(x)h.
However, in physics we still can say it, iff the error is less than say 2%. In other applied sciences that threshold may be higher (usually 5% in statistics).