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# How Much Work Must Be Done To Pull The Plates Apart To Where The Distance Between Them Is 2.0 Mm?  I Just got a question from my colleague related to his essay. The question is: How Much Work Must Be Done To Pull The Plates Apart To Where The Distance Between Them Is 2.0 Mm?

I’m really not sure about the answers, so I take a look at the browsers and doing some research related to the question. Here are my research results.

The original question is: A capacitor consists of two 6.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 150 V, then the battery is removed.

1. How much energy is stored in the capacitor?
2. How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?
• Capacitance (C) = (ε_o A)/d ; U=1/2*CV^2 ;
• C= (8.85*〖10〗^(-12)*π*(4.7/2*〖10〗^(-2) )^2)/〖10〗^(-3) = 1.535e-11.

For the second part, I am using these steps :.

• V_f=(170V*2mm)/1mm = 340 V.
• W= U_f-U_i= 8.874e-7J – 2.218e-7J= 6.656e-7.

• V_f=(170V*2mm)/1mm = 340 V.

After the plates are moved the capacitance changes:.

• Capacitance (Cn) = (ε_o A)/d ; U=1/2*CV^2 ;
• Cn= (8.85*〖10〗^(-12)*π*(4.7/2*〖10〗^(-2) )^2)/ 2 *〖10〗^(-3).

Determine the magnitude of the force on an electron traveling 6.35×10^5 m/s horizontally to the east in a vertically upward magnetic field?

Find the direction of the force on a negative charge for each diagram shown in the figure, where v⃗ (green) is the velocity of the…?

An isolated parallel-plate capacitor (not connected to a battery) has a charge of Q=1.4×10^-5 C. The separation between the plates initially is d=1.2 mm, and for this separation the capacitance is 3.1×10^-11 F.

Calculate the work that must be don to pull the plates apart until their separation becomes 4.5 mm if the charge on the plates remains constant.

The capacitor plates are in a vacuum. Related Calculus and Beyond Homework Help News on Phys.org. 618. You will probably also need a formula for the relation between the capacitance between the two plates and the distance between them, right?

Then use the potential energy equation. And leave the CAPS LOCK off, ok? 0. ok i have been tryin to figure this out for the past hour and got a work of 4.7×10^-10 J would that be correct? 618. Your capacitance is the only thing that changes.

• C_after=C_before*(1.2/4.5).
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Both of your energy numbers seem somewhat off to me. Isn’t (1/2)*(1.4E-5*coulomb)^2/(3.1E-11*farad)=3.16J?

You don’t have to convert the distance to any particular units. The capacitance is proportional to 1/D. You can’t directly use the e0*A/D formula because you don’t know A. But you do know A is the same at both distances. Take a ratio. 0.

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D cuz of the same reason we have no area which is why i used.

• C=(C*E)/(d)… To find capacitance at 4.5 mm…..

Because they gave us capacitance at 1.5 mm in the problem.

• U=(1/2)(Q^2)/C)… to find potential energy at 1.5 mm and at 4.5 mm.
• W=U2-U1… to find the work done.

Is what I did incorrect??, because we do not get the same answers…. Idk i may be wrong but what your doing seems like your just solving for capacitance after and not for the work.

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D cuz of the same reason we have no area which is why i used.

• C=(C*E)/(d)… To find capacitance at 4.5 mm…..

Because they gave us capacitance at 1.5 mm in the problem.

• U=(1/2)(Q^2)/C)… to find potential energy at 1.5 mm and at 4.5 mm.
• W=U2-U1… to find the work done. Is what I did incorrect?

If a capacitor is charged by a source with constant voltage, $V$, the energy, $W$, stored by a capacitor of capacitance $C$ is given by:. $$• W = \frac{1}{2} QV = \frac{1}{2} V^{2}C = \frac{Q^{2}}{2C}$$.

Where $Q$ is the charge stored on the plates. If the capacitor is now disconnected, and its capacitance is changed by increasing the distance between the plates, its energy must increase since work is being done by moving charge under the influence of an electrical field.

Furthermore, since the plates are not connected to anything, it can be assumed that charge is conserved. The method for calculating the change in energy that I have seen in textbooks and on the internet is to calculate $Q$ for the pre-separated case, calculate $C$ for both cases, then use:. $$• W = \frac{Q^{2}}{2C}$$. to give $$\delta W = \frac{Q^{2}}{2} \left( \frac{1}{C_{2}} – \frac{1}{C_{1}} \right)$$. where $C_{2}$ is capacitance post-separation, $C_{1}$ is capacitance pre-separation and $\delta W$ is the change in energy.

This seems naive to me. That formula was for the process of charging a capacitor from a constant voltage with the $\frac{1}{2}$ arising from the potential difference that charge is raised by decreasing as more charge is deposited. where $V_{2}$ is the post-separation voltage and $V_{1}$ is the pre-separation voltage. Substituting $Q = CV$ then gives:. $\begingroup$ In your integration you’ve assumed $Q$ is constant, but you know it’s a function of the potential difference,

• $Q=CV$. So $dW=CV\,dV$.

For simplicity, assume that the plate with charge $-Q$ is fixed and that the plate with charge $Q$ is movable. Making the usual simplifying assumptions for parallel plate capacitors, note that the electric field, due to the fixed plate, at the location of the movable plate is. $$E_\mathrm{fixed} = -\frac{Q}{2\epsilon A}$$. where $A$ is the area of a plate.

It follows that the work done slowly moving the plate a small distance is. $$\Delta W = F\delta d = (- Q E_\mathrm{fixed}) \Delta d = \frac{Q^2}{2\epsilon A} \Delta d = \frac{Q^2}{2}\left(\frac{1}{C_f} – \frac{1}{C_i}\right)$$. $\begingroup$ Since this demonstrates that all the work done in moving the plate ends up in the capacitor, my premise that it was possible to consider the charge on the plate as having moved from one potential to another and hence calculate an energy change was overly simplistic.

I assume that this is because $W=QV$ is based on a test charge changing position in an electrical field rather than the charge causing the electrical field changing position. I think I need to do some more reading.

You need to do work when you want to seperate two plate. Assuming the area of plate is S and distance between the plates is d(also assume $S\gg d^2$). Then $C=\frac {S}{\epsilon_0 d}$ in vacuum. The area charge density is $Q \over S$ and it makes electric field.

• $$E=\frac {Q}{2\epsilon_0 S}$$

So there is attractive force between the plates and you need to do work. Difference between two calculation equals to the amount of work.

As I understand it, then, moving the charge, Q, up a potential gradient through the mechanism of decreasing capacitance from C1 to C2 requires an amount of work according to my last equation. Half of this ends up in the electrical field, as would be calculated for a black box capacitor of capacitance C2 subject to a PD of Q/C2.

The other half ends up in the physical position of Q relative to its own electrical field. On further rumination, I’m not sure that this answers the question.

The work done in separating the plates is the source of the extra energy manifested in the increased voltage, it is not a sink for that energy. A capacitor with charge Q and capacitance C2 stores the same energy regardless of physical configuration or the path it took to that situation.

A potential of 100V is applied across the plates, which are. cm apart, using a storage battery. What is the energy stored in the capacitor? Suppose that the battery is disconnected, and the plates are moved until they are. cm apart. What now is the energy stored in the capacitor? Suppose, instead, that the battery is left connected, and the plates are again moved until they are. . Answer: The initial energy stored in the capacitor is. When the spacing between the plates is doubled, the capacitance of the capacitor is halved to.

Thus, it follows from the formula. that the energy stored in the capacitor doubles.

So, the new energy is. J. Incidentally, the increased energy of the capacitor is accounted for by the work done in pulling the capacitor plates apart (since these plates are oppositely charged, they attract one another). If the battery is left connected, then the capacitance is still halved, but now the process takes place at constant voltage.

It follows from the formula. that the energy stored in the capacitor is halved. So, the new energy is. J. Incidentally, the energy lost by the capacitor is given to the battery (in effect, it goes to re-charging the battery). Likewise, the work done in pulling the plates apart is also given to the battery.

I hope the explanation is good enough for you to answerring the question. I’m going to research for more answer to find the exact answer.