dielectricelectrostaticshomework-and-exercises
I am given two charged particles of same charge at a distance of $r$. They initially apply force $F$.
Now an infinite dielectric (of dielectric constant $4$) of width $\frac{r}{2}$ is introduced between the particles. What will be the new force?
I find this problem confusing because I have only been told about forces when its either fully dielectric or fully vaccum given by Coulomb's law. How do we get forces when only partial space is dielectric?
The attraction is that between a charge and an induced dipole. If the charged object is a sphere, the field is $QE\over 4\pi r^2$, in units where $\epsilon_0=1$. This field is what induces the dipole and causes the attraction.
Once you get a dipole moment in the dust, the dust dipole is attracted to regions of stronger field with a force that goes like the derivative of the E field. The total dipole is neutral, so you get net opposite positive and negative forces which only fail to cancel out to the extent that the E field is stronger in the regions of induced positive charge. The magnitude of the force is (in the nearly perfect small-dust approximation) $ d \cdot \nabla E$, the dipole moment can be thought of as defined by this equation (although that's not the definition--- the dipole moment vector d is the sum of qx over all the little infinitesimal regions in the dust, where q is the charge of the region, which sums to zero over the dust particle, and x is the position).
So if the dipole moment is of fixed size, the force is $2Qd\over r^3$. But the dipole itself is proportional to E, with a coefficient that I'll call "p" for polarizability, so you get a force which is
$$ {2 Q \over 4\pi r^3} { pQ\over 4\pi r^2} = {2pQ^2\over 16\pi^2 r^5}$$
and this is the force between a sphere and a dipole (in units where $k={1\over 4\pi\epsilon_0} = 1$). You see it falls off as $1/r^5$, two powers of r for the induced dipole strength, proportional to the E field, and another three powers from the force, which is as the gradient of E.
To calculate p requires knowing something about the material, namely it's dielectric polarizability constant for DC electric fields. This is radically different for different materials, depending on whether the molecules are polar themselves, and how mobile they are, whether it's liquid or solid. In general, you can figure this out from the extent to which the electric field in the interior of the dust is reduced from the exterior, assuming a bulk block of dust material. This is the dielectric constant $\epsilon$ of the dust.
For a solid block of dust in a constant electric field E perpendicular to the plane surface of the dust-block, the electric field in the interior of the dust block is $E\over \epsilon$ where $\epsilon$ for DC fields is always bigger than 1. For a metal, the field in the interior is zero, and this corresponds to infinite $\epsilon$. I will calculate the polarization constant in an order of magnitude for metal dust, but for the largely nonpolar typical carbon-chain dust material, the actual answer will range from 1%-10% of the metal answer.
For the metal, there is a nice useful trick for getting the induced dipole moment magnitude, which is the method of images/conformal maps. A dipole at the origin is conformally equivalent by an inversion around a sphere of certain radius to a constant electric field at infinity, so if you place a dipole at the origin, and the equal electric field inverted, the result makes the sphere have the same potential.
The electric potential of a dipole is by differentiating the electric potential of a point charge:
$$ \phi = {dz \over 4\pi r^3}$$
The potential of constant z-direction electric field is
$$ \phi = Ez$$
adding the two, the surface at potential zero (only a zero potential stays an equipotential under inversion) is a sphere:
$$ Ez - {dz\over 4\pi r^3} = 0$$
This is solved by a sphere (away from the z=0 plane which is an accidental equipotential by symmetry)
$$ d = 4\pi E r^3 $$
and so the induced dipole moment is proportional to the cube of the sphere size. The coefficient p is $4\pi a^3$ for a metal spherical dust of radius a.
So the answer for an ideal resistive metal dipole (perfectly screening object, zero response time to adjust to a new field, these are good assumptions for this situation) is
$$ F = {Q^2 a^3 \over 4\pi \epsilon_0 r^5} $$
Where I restored $\epsilon_0$ in the final answer. For a non-metal resistive dipole, the induced moment is less by the facor $(1-{1\over \epsilon})$. For hydrocarbons, the dielectric constant is about 2, so you get 1/2 the force. Note than when r is of order a, the maximum force goes up as the object gets small.
It's intuitive to me that if capacitor 1 is initially charged to 5 volts, for example, charge would be transferred until capacitor 2 has 2.5 volts across it, meaning that capacitor 1 is now charged to only 2.5 volts as well.
That would only be correct if the two capacitors had the same capacitance. If they are not the final voltage will not be half the initial voltage of the one capacitor. You can determine the final voltage as a function of the capacitances and initial voltage from three equations, (1) $q_{1}+q_{2}=q_0$, (2) $q_{0}=C_{1}V_0$, and (3) $\frac{q_1}{C_1}=\frac{q_2}{C_2}=V_{final}$.
I understand that without a potential difference, charge doesn't flow...
That is correct.
but I previously thought that this was because there was enough force from the accumulated charge at the plate of one capacitor to cancel out forces from the accumulated charge on the other capacitor plate.
When the switch is closed in the circuit depicted, charge redistributes itself until equilibrium is reached and the voltage on each capacitor is the same. Since $q=CV$, if the capacitances are different, the charge on each capacitor will necessarily be different. They are, however, in equilibrium and there is not net field between the capacitors as discussed below.
Do the authors intend to say that there is no NET electric field to influence the movement of charge?
I don't know what the authors intended to say, but it is true that after the switch is closed and charge redistributes itself so that the voltage is the same on each capacitor, there is then no net electric field to influence the movement of charge.
Wouldn't the positive charge on both plates of the capacitors still create an electric field, just equal electric fields whose forces cancel each other out?
See FIG A below. Although the positive charge on both plates of the capacitors creates an electric field pointing outward from each positive plate, the negative charge on both plates of the capacitors also creates an electric field of equal magnitude pointing inward to the positive plate, so that the net electric field outside the positive plate is zero.
Similarly, the electric fields pointing inward and outward on the negative plates of the capacitors also cancel so that the net electric field outside the negative plate is also zero.
The fact that the electric field outside the capacitor due to the charges on the plates is zero, can also be explained by Gauss' Law. See FIG B. Gauss' law states that the net electric flux across a closed surface equals the net charge enclosed by that surface divided by the permittivity of the space. Since the net charge on the overall capacitor is zero, the charge does not create an electric field outside the capacitor.
I'm not bridging the gap between potential differences and electric fields.
The relationship between the potential difference and electric field between two points is the gradient of the electrical potential between the two points. The electric field strength between the plates of a capacitor separated by distance $d$ and where the field is considered constant is then.
$$E=\frac{V}{d}$$
Since the plates of two parallel capacitors are at the same potential the electric field is zero between the plates of the parallel capacitors is also zero.
I've always understood that if a net charge exists and there isn't an equal net charge that produces a field that cancels its force out, there will still be a net electric field to influence charge to move.
Your understanding is correct, but as discussed above and in the figures below, in the case of a capacitor you can't look at the net charges on a plate in isolation from the other plate. While there is a net positive charge on one plate of the capacitor, there is an equal amount of net negative charge on the other plate so that, overall, the net charge on the complete capacitor is zero.
Hope this helps.
Best Answer
A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$.
To see this let's take the example we know about where the dielectric fills the space between the charges:
In (a) the thickness of the dielectric is the same as the distance between the charges, $r$, so the effective thickness is $r\sqrt{K}$. If we put this in the force law we get:
$$ F = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{(r\sqrt{K})^2} = \frac{1}{4\pi\epsilon_0}\frac{1}{K}\frac{Q_1Q_2}{r^2} $$
as we expect. The force is reduced by a factor of $K$.
Now consider (b). To get the effective distance between the charges we have to add the distance through the air, $r - t$, plus the effective thickness of the dielectric, $t\sqrt{K}$, so the effective distance between the charges is:
$$ d = (r - t) + t\sqrt{K} $$
and the force is just:
$$ F = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{d^2} = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{(r - t + t\sqrt{K})^2} $$
This is how you get the force when the space between the charges is only partially filled by the dielectric.