# A slab of glass dielectric is inserted into a parallel plate capacitor

• hidemi
In summary, the equations used to understand this question/answer are C = k(ε*Area)/distance = Q/V = Q/(E*distance) and F = QE. As a slab of glass is added, k increases and E decreases, causing the force to also decrease. This relates to the idea that the force attracts the glass into the capacitor. The potential energy stored in the capacitor can be represented by the equation F=-dU/dx, where U is the potential energy, C is the capacitance, k is the dielectric constant, ε is the electric field, Area is the area of the plates, and distance is the distance between the plates. By pushing the dielectric further between the plates, the potential energy decreases

#### hidemi

Homework Statement
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the
plates. As it is being inserted:
A. a force repels the glass out of the capacitor
B. a force attracts the glass into the capacitor
C. no force acts on the glass
D. a net charge appears on the glass
E. the glass makes the plates repel each other

Relevant Equations
C = k(ε*Area)/distance = Q/V = Q/ (E*distance)
F = QE
I use the following equations to understand this question/answer.
First, C = k(ε*Area)/distance = Q/V = Q/ (E*distance)
As a slab of glass is added, k increases and thus E decreases.

F=QE, as E decreases, force decreases as well. How does this relate to the 'force attracts the glass into the capacitor'

Let me know if my thoughts/logic on this is right. In addition, is there a less physic-based explanation? Thanks.

F = QE is not much help here. The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself.

I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant.

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• hidemi
kuruman said:
F = QE is not much help here. The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself.

I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant.
I used your equation, F=-dU/dx, and substitute in relevant symbols as followed:
U = CV/2 = 1/2* kεA/d *E*d = kεAE/2
Therefore, the force has nothing to do with the distance? How can I know if it is attracted or repeled?

You have the wrong ##x##. It is not the plate separation, it is the extent to which the dielectric is inserted between the plates. The derivative is with respect to that distance. The idea is to find the capacitance as a function of ##x## and then take the derivative. I have a feeling that this might be too complicated for you and you really don't need to do it that way.

You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that.

• hidemi
You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that.
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The potential energy is decreasing if I push the dielectric some more in between the plates.