The energy lost in discharging a capacitor

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The energy stored in the capacitor is $\frac { CV^2 } { 2 }$. So, this much energy will dissipate as heat. Hence, the correct option is (a).

Is this correct?

It doesn't depend on the shape of the capacitor, right?
I have done this calcculation assuming that the capacitor is parallel plate capacitor.

 

[lekh2003]

It doesn't depend on the shape of the capacitor, right?
I have done this calcculation assuming that the capacitor is parallel plate capacitor.

It probably is a parallel plate capacitor. I doubt that they would want you to assume its anything else in such a simple question.

 

[scottdave]

The shape of the capacitor, as well as the material used for dielectric, will determine the value of C.
The amount of energy stored in a capacitor will be the amount available to convert to other forms of energy (heat, light, etc.)
No energy is "lost", just converted to another type.

 

[lekh2003]

No energy is "lost", just converted to another type.

I think what OP means by "lost" is the wastage of energy. When speaking of machine efficiency and especially light-bulb efficiency, thermal energy is for all practical purposes, "lost" energy. This is because heat is seldom the aim of the light bulb or machine (of course this doesn't apply for heaters).

But, I do understand what you are saying.

 

[Pushoam]

It probably is a parallel plate capacitor. I doubt that they would want you to assume its anything else in such a simple question.

Even if it were not a parallel plate capacitor, will the above answer be correct?

 

[lekh2003]

Even if it were not a parallel plate capacitor, will the above answer be correct?

I would assume so. As already stated by Scottdave.

 

[scottdave]

Even if it were not a parallel plate capacitor, will the above answer be correct?

The shape and size and dielectric material will all go into determining the capacitance value. The energy stored in a capacitor is (½ CV²).
In the real world, the metal making up the capacitor and the wires will have a small resistance value which would turn some small amount of the electrical energy into heat.

 

[Pushoam]

Thanks for clarifying it.
The properties of material will determine C. After knowing C, the energy stored in the capacitor is ½CV².

The work done in bringing a charge dq to the capacitor positive plate from infinity is dW = Vdq, where V is function of q and q increases with time.
So, bringing charge Q, W = $\int_0 ^Q \frac q C dq = ½ \frac {Q^2}C = ½ CV^2$.

But, then the same amount of work is done in bringing the negative charge, too.
Yes, it is so.
Let's say that the potential difference between positive plate and negative plate is V. The positive plate has potential ½V and the negative plate has potential -½V wrt a point at infinity.

Then, the work done in bringing charges to the plates is W = ¼CV².

The above derivation is wrong.

We don't bring the charge from infinity to the positive plate.
We bring electrons from one plate to the other. The plate lacking electrons gets positively charged and the plate getting electrons gets negatively charged.
And the potential difference between them is V. So, the work done in bringing charge dq from one plate to the other is V dq, where V is a function of q, V = q/C.

So, the work done in bringing charge Q from one plate to another is W = $\int_0 ^Q \frac q C dq = ½ Q^2/C = ½ CV^2$.

So, the expression for the energy stored in the capacitor is always ½ CV².

Is this correct?

 

[scottdave]

Thanks for clarifying it.
The properties of material will determine C. After knowing C, the energy stored in the capacitor is ½CV²...

So, the expression for the energy stored in the capacitor is always ½ CV².

Is this correct?

Yes.

I didn't quote the middle part of your post, but it seemed almost as if that was a conversation going back between 2 people.

 

[Pushoam]

if that was a conversation going back between 2 people.

The conversation was going into my mind. This happens whenever I have to drive the energy stored in the capacitor. So, I posted it as it was going in my mind hoping that it will get verified by you and if there is anything wrong in it you may say it.

Thanks.

 

[Pushoam]

The above derivation is wrong.

I think the above derivation is correct if I bring positive and negative charge separately to both the plates.

To take the above derivation as the derivation of energy stored in a capacitor is wrong.

Is this correct?

But then how is it that the work done in bringing positive and negative charges separately to the correesponding plates is less( so less costlier ) than bringing only electrons from one plate to another (charging the capacitor)?

I think that it should be opposite. When I am bringing the positive and negative charges separately, I am increasing the charge density of the plates, too.

 

[scottdave]

If you bring excess electrons (negative charge) to one side of the capacitor, then the excess negative charge will act on free electrons on the opposite capacitor plate. So they will "want" to move into the wire away from the capacitor. That leaves a net positive charge on that end of the capacitor. The positive charge on one plate will always equal (in magnitude) to the negative charge on the other plate.

 

[Pushoam]

If you bring excess electrons (negative charge) to one side of the capacitor, then the excess negative charge will act on free electrons on the opposite capacitor plate. So they will "want" to move into the wire away from the capacitor. That leaves a net positive charge on that end of the capacitor. The positive charge on one plate will always equal (in magnitude) to the negative charge on the other plate.

I did not understand what you said. Sorry for this.
Could you please reformulate it?

When I am bringing charges from infinity, then in this case, I plates are not connected through wire.

The two plates are charged independently of each other, means the charging of one plate does not affect the changing of other plate.

In this way, the work done on the system is $\frac { CV^2} 4$.

But then , to arrange as a capacitor, I will have to bring the two plates at a distance d. In fact, the plates will come on their own as they are oppositely charged. I need to do work on keeping them separated at a distance d. I don’t need to do any work in keeping them separate. When they are at distance d, I have to hold them so that they do not move further. I am applying force, but since there is no displacement, there is no work done.

Thus, if I charged the two plates oppositely keeping them very far from each other from taking charges from infinity and then let them to come at a separation d, so that I will have a charged capacitor, then the work done in changing the capacitor this way is $\frac {CV^2}4$.

Is this correct?

 

[scottdave]

In fact, the plates will come on their own as they are oppositely charged. I need to do work on keeping them separated at a distance d. I don’t need to do any work in keeping them separate. When they are at distance d, I have to hold them so that they do not move further. I am applying force, but since there is no displacement, there is no work done.

While no work is required to keep them at a specific distance, you will have work while the force is applied decelerating the plates to a stop. It is not as if you can let them attract each other then apply a force and stop them instantly.

 

[scottdave]

If the plates are so far apart that they do not interact with each other, then really it is not a capacitor, hence the value of C in your formula $\frac { CV^2} 4$ has no meaning.

 

[Pushoam]

If the plates are so far apart that they do not interact with each other, then really it is not a capacitor, hence the value of C in your formula $\frac { CV^2} 4$ has no meaning.

The following part was not clear. In this part, I am charging the capacitor by keeping the plates very far and still applying V = q/C, which is wrong.

The two plates are charged independently of each other, means the charging of one plate does not affect the changing of other plate.

In this way, the work done on the system is $\frac { CV^2} 4$.

But then , to arrange as a capacitor, I will have to bring the two plates at a distance d.

Charging the capacitor by taking charges from infinity

A) Bringing both positive charges and negtive charges together simultaneously.

This way the charges on both plates kept at distance d will be always equal and opposite.

Potential difference between both plates , V = q/C. So, potential difference between one plate and infinity will always have magnitude V/2 = q/2C.

So, if I charge the capacitor this way, then the work done will be $\frac 1 4 CV^2.$B) But if I keep the plates very far and then I bring charges upon them separately, then the work done in bringing the charges on both plates will be $2\int_{ 0}^{ Q} V dq$ . Here, V = V(q), but unknown to me. V(q) $\neq q/C$ .

So, I do not know how to calculate work done in this case.

And then some work will be done in bringing the plates at a distance d.

Is this correct?
But, does it mean that when I have brought the plates at a distance d, the total work done on the capacitor is $½ CV^2$?

The energy stored in the capacitor will always be $½ CV^2$, irrespective of how I charged it as this energy is potential energy so it should be independent of the process of charging.

But, does the total work done on the capacitor depend on the way we charge it?
And in text-book question of calculating total work done in charging a capacitor,it is assumed that we are charging it by taking electrons from one plate to another plate?

 

[scottdave]

Think about this: what is the work required to move a charge from infinity along the wire? In the wire, recall that if you take an electron away from an atom, the atom now has a net positive charge and there will be an attractive force between charged atom and electron.