Work and Energy Loss; Electrostatics

But in terms of the equations, it does make a difference.In summary, the conversation discusses Griffith's Problem 2.40 (a) and (b), which involves calculating the work done by electrostatic forces and the energy lost by the field in a parallel-plate capacitor as the plates move closer together. The problem requires the answer to be in terms of the area of the plates, A, and the electric field, E. The person discussing their solution notes that they expanded their E value, but Griffiths' solution does not include this. The conversation also considers whether the potential difference across the plates was kept constant, as this would affect the calculations.
  • #1
mateomy
307
0
Griffith's Problem 2.40 (a) and (b)

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance ε, as a result of their mutual attraction.

a) Use

[tex]
P= \frac{\epsilon_0}{2}E^2
[/tex]

to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.

b) Use

[tex]
\frac{\epsilon_0}{2}E^2 = energy per unit volume
[/tex]

to express the energy lost by the field in this process.I solved the problems in my (they're both the same answer in the answer key) way getting

[tex]
\frac{(q^2)\epsilon}{2A\epsilon_0}
[/tex]

But Griffiths doesn't do that, he doesn't even delve into what E is, which I solved ambiguously using A. His answer was:

[tex]
\frac{\epsilon_0}{2}(E^2)A\epsilon
[/tex]

Is it alright that I expanded my E?
 
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  • #2
mateomy said:
Griffith's Problem 2.40 (a) and (b)

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance ε, as a result of their mutual attraction.

a) Use

[tex]
P= \frac{\epsilon_0}{2}E^2
[/tex]

to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.

b) Use

[tex]
\frac{\epsilon_0}{2}E^2 = energy per unit volume
[/tex]

to express the energy lost by the field in this process.


I solved the problems in my (they're both the same answer in the answer key) way getting

[tex]
\frac{(q^2)\epsilon}{2A\epsilon_0}
[/tex]

But Griffiths doesn't do that, he doesn't even delve into what E is, which I solved ambiguously using A. His answer was:

[tex]
\frac{\epsilon_0}{2}(E^2)A\epsilon
[/tex]

Is it alright that I expanded my E?

Well, not really. The poblem demands that the answer be in terms of A and E.

However: was it specified whether the potential difference across the plates was kept constant, or was the voltage source used to charge the capacitor removed before the displacement took place? Makes a big difference ...
 
  • #3
It didn't specify. What I posted was as much as he wrote in for the particular question. But the point you brought about about A and E was overlooked and that makes sense.
 
  • #4
rude man said:
However: was it specified whether the potential difference across the plates was kept constant, or was the voltage source used to charge the capacitor removed before the displacement took place? Makes a big difference ...

No, it makes a small difference (infinitesimally small, to be precise ;-P).
 
  • #5
Steely Dan said:
No, it makes a small difference (infinitesimally small, to be precise ;-P).

Well, if you don't consider 2:1 as big, then OK, it makes no big difference.
 

1. What is work and energy loss?

Work and energy loss refers to the amount of energy that is expended or lost when a force is applied to an object and causes it to move. This can be due to factors such as friction, air resistance, or other forms of resistance that oppose the movement of the object.

2. How is work and energy loss related to electrostatics?

Electrostatics is the study of electric charges and their interactions. When an object with an electric charge moves, work is done and energy is lost due to the electrical forces involved. This can occur in situations such as charging a capacitor or moving electrically charged particles through a circuit.

3. What factors affect work and energy loss in electrostatics?

The amount of work and energy loss in electrostatics is affected by the strength of the electric field, the distance the object moves, and the amount of charge on the object. The type of material the object is moving through can also impact the amount of work and energy loss.

4. How can work and energy loss be minimized in electrostatic systems?

To minimize work and energy loss in electrostatic systems, it is important to reduce the amount of friction or resistance that the object experiences. This can be achieved by using materials with low electrical resistance, lubricating surfaces, and minimizing the distance the object needs to travel.

5. Can work and energy loss be completely eliminated in electrostatics?

No, it is not possible to completely eliminate work and energy loss in electrostatic systems. This is because there will always be some resistance or friction present, even in the most optimized systems. However, it is possible to minimize work and energy loss to a very low level through careful design and optimization of the system.

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