How to Derive the Expression for \(\delta W = -E dP\)?

[appmathstudent]

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.

 

[anuttarasammyak]

$$dP=d(qD)=(dq) D+ q (dD)$$
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 ?

 

[appmathstudent]

I think q is changing, since the work is done to change E.

 

[bob012345]

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.

 

[anuttarasammyak]

I think q is changing, since the work is done to change E.

q is changing, d is constant so
$$dP=d(dq)$$
$$\frac{dW}{dP}=\frac{1}{d}\frac{dW}{dq}$$
Try it.