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

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

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closed as off-topic by AccidentalFourierTransform, Kyle Kanos, Yashas, John Rennie, ZeroTheHero Jul 31 '18 at 2:37

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I am supposed to get $k^2 F$. What am I missing?

You actually added something that should not be added - the permittivity of dielectric in the force formula $$ F = \frac{q^2}{2\epsilon A}. $$

If the dielectric is a solid slab inserted in between the plates, this is not the correct formula. The correct formula is

$$F = \frac{q^2}{2\epsilon_0 A}. $$

The reason is that the field acting on the capacitor plate is entirely due to the other capacitor plate; the field due to dielectric is zero outside the dielectric.

If the dielectric is a fluid filling the space in between and outside the plates, things change: the fluid exerts some pressure on the plates and the formula (1) becomes correct (this can be derived using the principle of virtual work).

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The force $F$ between capacitor plates is discussed here and shown to be $\frac 12 QE$ where $Q$ is the charge on the capacitor and $E$ the electric field strength.

In your example with a constant voltage this is better written as $F=\frac 12 CV \; \frac V d $ where $C$ is the capacitance $=\frac{k\epsilon_o A}{d}$ with $d$ the separation of the plates and $A$ the area of the plates.

So the force $\left (=\frac {k\epsilon_o A V^2}{d^2}\right )$ is proportional to the dielectric constant $k$.

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  • $\begingroup$ The referenced derivation is for vacuum-separated plates and $E$ is the field in the vacuum. If there is a solid dielectric between the plates, the result is still correct, but $E$ must be taken to be the field outside the dielectric, just above the plate. In this question, that field is $E=kV/d$ (field inside the dielectric is $V/d$ but is $k$ times weaker due to depolarization field of the dielectric). So the force increases $k^2$ times. $\endgroup$ – Ján Lalinský Jun 19 '18 at 9:37
  • $\begingroup$ @JánLalinský In the light of your comment should I correct my answer or just delete it? $\endgroup$ – Farcher Jun 19 '18 at 9:44
  • $\begingroup$ I think it is best if you correct it, perhaps you can find a better formulation than I did. $\endgroup$ – Ján Lalinský Jun 19 '18 at 9:50
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The electric field close to one of the plates (say the positive plate) is going to be $E_+= \frac{\sigma}{2\epsilon_o}=\frac{q}{2\epsilon_oA}$, which will generate a force on the other plate of $F=qE_+=\frac{q^2}{2\epsilon_oA}$. The field between the plates is going to be $E=V/d$ (where $d$ is the distance between the plates) and the charge on each plate is going to be $q=CV$. These are the same for a parallel plate capacitor in vacuum or with a dielectric.

Introducing a dielectric will increase the capacitance from $C_o$ (the capacitance without dielectric) to $C$ by a factor of $k$, and therefore the amount of charge on each plate by $k$. This is because $C_o=\frac{\epsilon_oA}{d}$ becomes $C=\frac{k\epsilon_oA}{d}=kC_o$, and therefore $q=CV=kC_oV$.

Putting all this together the resulting force is $F=\frac{k^2\epsilon_oA}{2d^2}V^2=k^2F_o$. Where $F_o$ is the force without a dielectric.

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