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I was told that a dielectric slab inserted into a capacitor connected to a battery (constant voltage) will be repelled, because the energy stored in the capacitor increases when the dielectric is inserted, due to increased capacitance. What is the physical origin of this force? The attractive force by a constant charge capacitor can be explained by fringe field, but why are the two cases different in terms of forces?

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A dielectric slab will be attracted towards any source of electric field, because it is essentially a collection of dipoles. Dipoles in fields align with the fields. If the field varies in space, the dipole is drawn to the local maximum of the field. An intuitive, everyday example of this is the way iron filings are drawn to a magnet.

In the case of a constant voltage capacitor, the battery supplies the extra energy. We can see this most simply in the case that we replace the battery with a very large capacitor of capacitance $C$ connected in parallel with our original capacitor with capacitance $c_0 \ll C$. Originally, each capacitor is at a voltage $V_0$, and the large capacitor has charge $Q_0=CV_0$ while the small capacitor has charge $q_0=c_0V_0$. As we insert the dielectric, $c_0$ changes by $\Delta c$. The capacitors then settle in the new condition of equal voltage

$$\frac{Q_0-\Delta q}{C} = \frac{q_0 + \Delta q}{c_0 + \Delta c}$$

Solving for $\Delta q$ using the fact that $c_0 +\Delta c \ll C$ gives

$$\Delta q = Q_0 \frac{c_0 + \Delta c}{C} - q_0$$

From this it is fairly easy to show that the energy change in the original capacitor is

$$\Delta U_c=\frac{1}{2}\Delta c V_0^2$$

What about the change in the large capacitor? There we have

$$\Delta U_C = \frac{1}{2C}((Q-\Delta q)^2 - Q^2) \approx -\frac{\Delta q\, Q}{C}$$

Substituting gives

$$\Delta U_C=-\Delta c V_0^2$$

So overall the system ends with a lower energy, and the slab should be attracted.

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  • $\begingroup$ I was told about a constant voltage capacitor, so wouldn't we use the formula $U=\frac{CV^2}{2}$? $\endgroup$ – Poon Levi Mar 7 '15 at 5:05
  • $\begingroup$ Sorry! I did not read the question carefully. I have edited to address the case you were actually asking about. $\endgroup$ – user27118 Mar 8 '15 at 4:02

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