Find spring constant k using an electrical circuit

[ananonanunes]

Homework Statement

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Relevant Equations

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This circuit consis of two identical, parallel metal plates free to move, other than being connected to identical metal springs, a switch, and a battery with terminal voltage $\Delta V$. With the switch open, the plates are uncharged, are separated by a distance $d$, and have a capacitance of $C$. When the switch is closed, the plates become charged and attract each other. The distance between the plates changes by a factor $f$, after which the plates are in equilibrium between the spring forces and the attractive electric force between the plates. I am supposed to find spring constant $k$ in terms of $C$, $d$, $\Delta V$ and $f$.

So what I did was equal the spring force to the electric force. The spring force I wrote as $F_s=kdf$ and the electric force as $F_e=qE=q\frac{\Delta V}{d(1-f)}=\frac{C(\Delta V)^2}{d(1-f)}$. This gives a final result of $k=\frac{C(\Delta V)^2}{d^2f(1-f)}$.

I know this is wrong, and that the correct answer is actually $k=\frac{C(\Delta V)^2}{d^2f^2(1-f)}$. Since my answer is so similar to the correct one, I feel like maybe I'm just missing something simple, but I can't seem to understand where that extra $f$ could possibly come from.

 

[kuruman]

With the switch open the capacitance is $C$. Is it still $C$ when the switch is closed?

 

[ananonanunes]

With the switch open the capacitance is $C$. Is it still $C$ when the switch is closed?

I think the capacitance depends on the distance between the plates, so no. Would this mean my mistake was assuming $q=C\Delta V$?

 

[kuruman]

I think the capacitance depends on the distance between the plates, so no. Would this mean my mistake was assuming $q=C\Delta V$?

No. $q=C\Delta V$ always holds because it is the definition of capacitance. Your mistake is in assuming that the capacitance does not change. The symbol $C$ in the correct answer is the capacitance before the switch is closed. You have to find the new capacitance in terms of the old capacitance and put that in your expression.

 

[ananonanunes]

No. $q=C\Delta V$ always holds because it is the definition of capacitance. Your mistake is in assuming that the capacitance does not change. The symbol $C$ in the correct answer is the capacitance before the switch is closed. You have to find the new capacitance in terms of the old capacitance and put that in your expression.

Okay so for a parallel plate capacitor I have $C=\frac{\epsilon_0 A}{d}$ so the final capacitance should be $C_f=\frac{\epsilon_0 A}{d(1-f)}=\frac{C}{1-f}$. If I write $q=\frac{C}{1-f}\Delta V$ I would get $k=\frac{C(\Delta V)^2}{d^2f(1-f)^2}$ which is still not the correct answer but makes me wonder whether I'm misunderstanding the meaning of $f$ and maybe I should use $fd$ for the "new distance" and $d(1-f)$ for the displacement instead?

 

[kuruman]

$\dots$ makes me wonder whether I'm misunderstanding the meaning of $f~\dots$

If the separation $d$ changes by a factor $f$, then the new separation is related to the old by $~d_{\text{new}}=f~d_{\text{old}}.$ "Factor" implies multiplication.

 

[ananonanunes]

If the separation $d$ changes by a factor $f$, then the new separation is related to the old by $~d_{\text{new}}=f~d_{\text{old}}.$ "Factor" implies multiplication.

Okay, that makes sense. Thank you so much for your help!

 

[kuruman]

You do not show the details of your calculation but if you got the correct answer, after fixing the misunderstanding about the factor $f$, then you probably made two mistakes that cancelled each other.

1. The electric force on one of the capacitor plates is not $F_e=qE.$ It is $F=\frac{1}{2}qE$. You can derive this expression by noting that, at constant voltage, the force is the negative derivative of the potential energy stored in the capacitor $$F_e=-\frac{d}{dx}\left(\frac{1}{2}CV^2\right)=-\frac{d}{dx}\left(\frac{1}{2}\frac{\epsilon_0A}{x}V^2\right)=\frac{1}{2}\frac{C}{x}V^2.$$2. The force so derived must be equal to the force exerted by one spring which is stretched by distance $\xi=\dfrac{d_{\text{old}}-d_{\text{new}}}{2}=\dfrac{1}{2}d(1-f).$

You can see that when you set the magnitudes equal, $F_e=k\xi$, the factors of $\frac{1}{2}$ cancel.

I posted this for your benefit and that of future readers who might surmise that your initial expression for the force is correct.

 

[ananonanunes]

You do not show the details of your calculation but if you got the correct answer, after fixing the misunderstanding about the factor $f$, then you probably made two mistakes that cancelled each other.

1. The electric force on one of the capacitor plates is not $F_e=qE.$ It is $F=\frac{1}{2}qE$. You can derive this expression by noting that, at constant voltage, the force is the negative derivative of the potential energy stored in the capacitor $$F_e=-\frac{d}{dx}\left(\frac{1}{2}CV^2\right)=-\frac{d}{dx}\left(\frac{1}{2}\frac{\epsilon_0A}{x}V^2\right)=\frac{1}{2}\frac{C}{x}V^2.$$2. The force so derived must be equal to the force exerted by one spring which is stretched by distance $\xi=\dfrac{d_{\text{old}}-d_{\text{new}}}{2}=\dfrac{1}{2}d(1-f).$

You can see that when you set the magnitudes equal, $F_e=k\xi$, the factors of $\frac{1}{2}$ cancel.

I posted this for your benefit and that of future readers who might surmise that your initial expression for the force is correct.

This is very helpflul, I would have never noticed the mistake by myself. Thanks for pointing it out

 

[TSny]

You can derive this expression by noting that, at constant voltage, the force is the negative derivative of the potential energy stored in the capacitor $$F_e=-\frac{d}{dx}\left(\frac{1}{2}CV^2\right)=-\frac{d}{dx}\left(\frac{1}

The plates attract each other, so each plate experiences a force that tries to decrease the plate separation $x$. But from energy principles, if a plate were to move in the direction of the force on the plate, then the potential energy of the system should decrease. From the expression $U_{\text{cap}} = \large \frac 1 2 \frac{\varepsilon_0 A}{x} \normalsize V^2$, we see that if $x$ decreases, then the potential energy of the capacitor increases!

This “paradox” is resolved by realizing that the potential energy of the system should include the energy $U_{\text{bat}}$ stored in the battery. The total potential energy of the system $U_{\text{tot}} = U_{\text{cap}} + U_{\text{bat}}$ can be shown to decrease if $x$ decreases (even though $U_{\text{cap}}$ increases).

In fact, you can show $$ \frac{dU_{\text{bat}}}{dx}= -2 \frac{dU_{\text{cap}}}{dx}.$$ So, $$\frac{dU_{\text{tot}}}{dx}= - \frac{dU_{\text{cap}}}{dx}$$.

Therefore, calculating the magnitude of the force on a plate using $|F_e| = |\large \frac {dU_\text{tot}}{dx}|$ gives the same result as using $|F_e| =|\large \frac {dU_\text{cap}}{dx}|$. Thus, $$ |F_e| = \frac 1 2 \frac{C}{x}V^2$$ as obtained in post #8.

 

[nasu]

This is just an example of the force exrted by an electric field on a surface charge. In general, the force per unit area is given by $f= \sigma E_{average}$ where $E_{average}$ is the average of the values of the field on the two sides of the surface. Here the charge is on the inner surface of the plate and the field on the other side is zero (inside the conducting plate). So, $f= \frac{1}{2} \sigma E$ where $E$ is the field between the plates. (here f is the force per unit area so the force on the plate will be $F=fA=\frac{1}{2} q E$ as indicated by @kuruman in post 8.