Showing that ##\delta W = -E dP##

  • Thread starter appmathstudent
  • Start date
In summary, the conversation discusses the work, W, in relation to a dielectric slab with a constant electric field, E. The question asks how to eliminate the 2 in the denominator of the expression for W. The answer is to use the equation dP=d(qD)=(dq) D+ q (dD), where D is the distance between the capacitor plates. The conversation concludes that q is changing during the charging process, and therefore dP=d(dq) and $\frac{dW}{dP}=\frac{1}{d}\frac{dW}{dq}$.
  • #1
appmathstudent
6
2
Homework Statement
This is exercise 3-12 form Sears and Salinger Thermodynamics : Show that $$\delta W = -E dP$$ by calculating the work necessary to charge a parallel plate capacitor containing a dielectric.
Relevant Equations
$$W = U = \frac{q^2}{2C}$$
$$ C = \frac{\kappa \epsilon_0 A}{d}$$
$$P = qd $$ (dipole moment of slab)
$$ E = \frac{q}{\epsilon_0 \kappa A}$$
$$W = U = \frac{q^2}{2C} =\frac{q q d}{2 \kappa \epsilon_0 A} = \frac{E P}{2}$$

Then , since E is constant we have that :

$$\delta W = \frac{dW}{dP} dP = \frac{E}{2} dP$$.

My question is how can I make this 2 on the denominator disappear in order to obtain the required expression ?

ps : In the book (Chapter 3 page 67) he mentions that $$\delta W$$ is the work when $$E$$ is changed in a dielectric slab.
 
Last edited:
Physics news on Phys.org
  • #2
[tex]dP=d(qD)=(dq) D+ q (dD)[/tex]
where D is distance between the capacitor plates introduced to distinguish it with differential d.
Which is your case changing charge or changing distance or the both ?
 
Last edited:
  • #3
I think q is changing, since the work is done to change E.
 

Attachments

  • 20211002_125815.jpg
    20211002_125815.jpg
    30 KB · Views: 96
  • #4
appmathstudent said:
I think q is changing, since the work is done to change E.
So you are saying E is not constant during the charging process.
 
  • #5
appmathstudent said:
I think q is changing, since the work is done to change E.
q is changing, d is constant so
[tex]dP=d(dq)[/tex]
[tex]\frac{dW}{dP}=\frac{1}{d}\frac{dW}{dq}[/tex]
Try it.
 
  • Like
Likes appmathstudent

1. What is the significance of showing that ##\delta W = -E dP##?

The equation ##\delta W = -E dP## is known as the work-energy theorem, which states that the work done on a system is equal to the change in its kinetic energy. This equation is important because it allows us to calculate the work done on a system without explicitly knowing the forces acting on it.

2. How can we prove that ##\delta W = -E dP##?

The proof of the work-energy theorem involves using the definition of work as the dot product of force and displacement, and the definition of kinetic energy as ##E = \frac{1}{2}mv^2##. By manipulating these equations, we can show that ##\delta W = -E dP##.

3. What are some real-world applications of the work-energy theorem?

The work-energy theorem has numerous applications in physics and engineering. It is used to analyze the motion of objects in various systems, such as collisions, pendulums, and springs. It is also used in the design of machines and structures, as it helps determine the amount of work needed to move an object or lift a load.

4. Are there any limitations to the work-energy theorem?

While the work-energy theorem is a useful tool for analyzing the motion of objects, it does have some limitations. It assumes that there are no non-conservative forces acting on the system, such as friction or air resistance. In reality, these forces can affect the work done on a system and may need to be taken into account for more accurate calculations.

5. How does the work-energy theorem relate to the law of conservation of energy?

The work-energy theorem is a specific application of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. The work done on a system is a form of energy transfer, and the work-energy theorem shows that this transfer is equal to the change in kinetic energy of the system.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
648
  • Advanced Physics Homework Help
Replies
3
Views
841
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
2K
  • Quantum Physics
Replies
15
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
286
  • Advanced Physics Homework Help
Replies
4
Views
4K
Back
Top