How much work would be required to remove a metal sheet from b...

Question

How much work would be required to remove a metal sheet from between the plates of a capacitor (as in Problem 18a), assuming the battery remains connected so the voltage remains constant?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, a parallel plate capacitor is connected to a battery that maintains a constant voltage of V. A dielectric slab with a dielectric constant Kappa is inserted between the plates. You need to remove the dielectric slab while the battery remains connected, how much work is required to remove the slab under these conditions. So that's our end goal. Our end goal, what we're ultimately trying to solve for is we're trying to figure out how much work will be required in order to remove these slabs given or remove the slab, just the one slab to remove this slab from this parallel plate capacitor that's connected to a battery. So we're trying to remove the slab and we're trying to figure out how much work it will take to do that given all the conditions provided to us by the problem itself. So ultimately, we're trying to figure out how much work is required to remove the slab and that is our final answer that we're ultimately trying to solve for. So now that we know what we're solving for, let's read off our multiple choice answers to see what our final answer might be. A is epsilon zero multiplied by a multiplied by V squared, all divided by two multiplied by D multiplied by one minus Kappa B is epsilon zero multiplied by a multiplied by V squared, all divided by two multiplied by D multiplied by Kappa C is epsilon zero multiplied by V squared multiplied by Kappa all divided by a multiplied by D. And finally, D is epsilon zero multiplied by a multiplied by V squared multiplied by Kappa all divided by two multiplied by D. OK. So first off, we need to understand the initial setup. So we need to note that we have a parallel plate capacitor with a dielectric slap that is inserted between the plates and the connected battery will keep the voltage V across the capacitor constant. So with that in mind, we now need to determine the initial capacitance which we're gonna call this or denote the initial capacitance as C I with the dielectric slab. So we need to determine the initial capacitance with the dielectric slab. So as we should recall, the initial capacitance with the dielectric slab, this equation can be written as C I is equal to Kappa multiplied by epsilon zero multiplied by A all divided by D Awesome where Kappa is the dielectric constant of the slab epsilon zero is the permittivity of free space. A is the area of the plates and D is the separation between the plates. So now our next step is we need to determine the final capacitance, which we're gonna denote as CF. So we need to determine the final capacitance without the dielectric slab. So as we should recall the final capacitance, when the dielectric slab is removed, we could write this equation as CF is equal to epsilon zero multiplied by a all divided by D. So now our next step is we need to calculate the initial energy stored in the capacitor with the dielectric slab. So as we should recall, the initial energy stored in the capacitor with the dielectric slab can be written as U I is equal to C multiplied by V squared, all divided by two and U I is equal to KA multiplied by epsilon zero, multiplied by a multiplied by V squared all divided by two, multiplied by D. Awesome. So now our next step is we need to calculate the final energy stored in the capacitor without the dielectric slap. So as we should be able to recall the final energy stored in the capacitor. Without the dielectric slab, we could write this equation as UF and UF is equal to CF multiplied by V squared all divided by two. We can also expand this to write that UF is equal to epsilon zero and epsilon zero is multiplied by a multiplied by V squared, all divided by two multiplied by D. So now our next step is we need to calculate the work required to remove the dielectric slab, which is a final answer that we're ultimately trying to solve for. So as we should recall, the work required to remove the dielectric slab is the change in energy of the capacitor. Meaning that we could write that the work W is equal to UF minus U I. So now all we have to do is plug in our known variables and substitute in U I and UF into our work equation to get that the work is equal to epsilon zero, multiplied by a multiplied by V squared all divided by two multiplied by D for UF And we need to subtract this from our second equation which is Kappa multiplied by. And this is our value for U I. So Kappa multiplied by epsilon zero multiplied by a multiplied by V squared all divided by two multiplied by D Awesome. So we can begin to simplify this equation to get that the work W is equal to epsilon zero multiplied by a multiplied by V squared, all divided by two multiplied by D multiplied by one minus Kappa Awesome. So now moving right along here, we need to note that sense, Kappa is greater than one. This will give a negative work value which will indicate that the work done on the system when the dielectric is removed. So we need to note that the work required to remove the dielectric slab from between the plates of the capacitor while keeping the voltage constant is epsilon. So W is equal to epsilon zero multiplied by a multiplied by V squared all divided by two multiplied by D multiplied by one minus Kappa, which we can also write this expression as how we see it written in multiple choice answer A which states that epsilon zero multiplied by a multiplied by V squared all divided by two, multiplied by D multiplied by one minus Kappa. Awesome. And that's it. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

How much work would be required to remove a metal sheet from between the plates of a capacitor (as in Problem 18a), assuming the battery remains connected so the voltage remains constant?

Key Concepts

Capacitance

Work Done on a Capacitor

Dielectric Constant

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