Force between plates of a capacitor with dielectric and cell

[cr7einstein]

Homework Statement

My teacher told me that the force will increase $k^2$ times, where $k$ is the dielectric constant, but I don't see how.

Homework Equations

The Attempt at a Solution

To start with, with no dielectric, the force between the plates is given by $\frac{q^2}{2A\epsilon_{0}}$. If I do insert a dielectric, and The plates are connected to a battery, the charge becomes $q'=kq$; as potential difference doesn't change (thanks to the battery) but capacitance does. So, when I replace $q$ by $q'$ in the above equation, I must also simultaneously change $\epsilon_{0}$ to $k \epsilon_{0}$( as the medium has changed), so I get

$F'=\frac{k^2q^2}{2Ak\epsilon_{0}}=kF$, as one of the $k$'s gets canceled out. I am supposed to get $k^2F$. What am I missing? Thanks in advance!

 

[TSny]

So, when I replace $q$ by $q'$ in the above equation, I must also simultaneously change $\epsilon_{0}$ to $k \epsilon_{0}$( as the medium has changed),

You might need to reconsider this statement. It helps to imagine a very thin space (vacuum) between each plate and the neighboring dielectric surface. The electric field at the surface of one of the plates is the electric field just outside the dielectric material.

 

[conscience]

Adding to TSny's excellent reply -

It would be better to go back to the basics . How do you get the expression for the force on a plate of a capacitor ? You get it by multiplying the " net Electric Field at the plate" with "charge on the plate" . You have correctly calculated the charge . Now think about the net electric field . Which charges are contributing to this field ? Do these charges change on insertion of dielectric ? In case they change , by what factor do they change ? This should clear your doubt .

 

[Delta2]

The surface of the plate is a point (actually a set of points) of left and right discontinuity of the electric field, from left because the electric field in the interior of the plate is zero, and from right because there is the dielectric material there.

 

[conscience]

and from right because there is the dielectric material there.

Please read post#2 . I do not understand how does your post directly address OP's doubt .

 

[Delta2]

Isn't the value of the electric field inside the dielectric $\frac{kq}{2Ak\epsilon_0}$ while in the surface of the plate we take it to be $\frac{kq}{2A\epsilon_0}$ so there is a discontinuity from the right too...

 

[rude man]

I would suggest the principle of virtual work: displace the plates by a small amount, compute the change in stored energy, then equate that to force times the displacement plus the energy furnished or removed by the battery.

The problem is the same whether the dielectric is air or some higher-k dielectric.