Work Done in moving the plates of a Capacitor

  • Thread starter zorro
  • Start date
  • #1
1,384
0

Homework Statement


Consider a parallel plate capacitor connected to a battery. You move the plates closer to each other. Will you do positive or negative work?

The Attempt at a Solution


As the plates are moved closer to each other, capacitance increases which increases the stored energy (V is constant). We know that electrostatic potential energy is defined as the negative of the work done by the electric field. Ufinal - Uinitial is positive in this case. So work done by field is negative. Therefore work done by external agent (us) should be positive.
BUT THE ANSWER IS NEGATIVE WORK!!! please help
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
The voltage is constant and the capacitance is increased. That means increased charge. Where does this charge come from?

ehild
 
  • #3
1,384
0
The extra charge comes from the work done by the person in moving the plates closer. The problem is to find out the sign of work done.
 
  • #4
ehild
Homework Helper
15,543
1,909
The charge does not come from the work of the person, but from the battery. The work changes the distribution of the charge.

ehild
 
  • #5
1,384
0
So the battery in supplying charges does positive work. Does it mean that the person does negative work?
If yes then what is wrong with the explanation I gave in attempt at solution.
 
  • #6
ehild
Homework Helper
15,543
1,909
. So work done by field is negative. Therefore work done by external agent (us) should be positive.
Why is the work of the field negative? What if the positive plate of the capacitor is released and let move towards the negative plate. Will not it move by itself? The electric fields points in the direction of displacement. So the work of the field is positive.


ehild
 
  • #7
1,384
0
If the plates move by themselves, we are not doing any work in bringing them closer.Are we?
May be we do some work to stop the plates from getting in contact with each other. Is it right?
 
  • #8
ehild
Homework Helper
15,543
1,909
Yes. But usually it is assumed that the person exerts so much force that the whole process is quasi-stationary, with no change of kinetic energy. And as the person exerts force opposite to the force of electric field, its work is negative.

ehild
 
  • #9
1,384
0
Thanks!
 
  • #10
ehild
Homework Helper
15,543
1,909
The change of the electric energy of the whole system consisting of the battery and capacitor is equal to the external work if there are no other kinds of energy involved as kinetic energy or internal energy (increase of temperature). The energy of the capacitor Ec really increases if the separation of the plates decreases. ΔEc=1/2 ΔC V2. At the same time, the charge increases on the plates by ΔQ=ΔC V. This charge is supplied by the battery, where the chemical forces do work when separate the positive and negative charges. This work of the battery is Wb=ΔQ V. The external work is the sum of that of the person and that of the battery. ΔEc=Wp+Wb, 1/2 ΔC V2=Wp+ΔC V2, Wp=-1/2 ΔC V2.

ehild
 
  • #11
1,384
0
Thanks!
That was much more clear.
 

Related Threads on Work Done in moving the plates of a Capacitor

Replies
1
Views
4K
Replies
7
Views
4K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
269
Replies
4
Views
5K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
4K
Top