If I insert a piece of glass between two objects carrying different charges, would they still attract?

If they attract, does the piece of glass affect the force of attraction and is there any formula explaining the decrease of this force ? Also If they don't attract, Why..?


3 Answers 3


As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity $\epsilon_r$ (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be $C=\frac{\epsilon_0\epsilon_rA}{d}$. Here $\epsilon_0=8.854×10^{-12}C^2N^{-1}m^{-2}$ and $\epsilon_r=1$ for air. In case if the dielectric occupies only some of the space between the objects and the remaining air, then the capacitance would be $$C=\frac{\epsilon_0A}{(d-t)+\frac{t}{\epsilon_r}}$$ where $d$ is the distance of separation and $t$ is the thickness of dielectric. As the capacitance has been determined, the energy stored in a capacitor, $U=\frac{q^2}{2C}$ and also energy, $U=F.d$ $$F=\frac{q^2}{2Cd}$$ $C$ is substituted depending upon the extent to which the medium is present. As $\epsilon_r>1$ for any dielectric other than air, the capacitance of a dielectric-filled capacitor is always greater than that of the air-filled capacitor. Implying that the above force is always less than the Coulombic force of attraction between the two charged bodies (but, always depending upon the dielectric)..!

BUT, there would always be some (negligible atleast) attraction between the two bodies depending upon the $\epsilon_r$ value. Also, If you require the Coulombic force in order to compare both, $$F=\frac{q^2}{4\pi\epsilon_0d}$$


Yes, however you will change the capacitance of the capacitor based on the dielectric properties of the glass you are using and the material you are replacing (which might be air or even vacuum).

The general force between the two plates can be calculated as:

$$F = \dfrac{1}{2}\dfrac{Q^2}{Cd}$$

Where Q is the charge, C is the capacitance and d is the distance between the plates. Capacitance can be related to permittivity $\varepsilon$ and Area A and distance d and the dielectric constant k:

$$C = \dfrac{k\varepsilon_0A}{d}$$

One can find that the dielectric constant for glass is 5 times more than vacuum, and since the k would appear in the denominator of the force equation, the force between the plates decreases as the dielectric constant increases. Meaning the force exerted when one places glass instead of vacuum between the plates decreases.

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    $\begingroup$ I think you are right if the gap is completely filled with glass. But what if the glass did not completely fill the space, in other words, if there were still air gaps on each side of the glass. Wouldn't the electric field in the air be the same with or without the glass and so wouldn't the force be the same? $\endgroup$
    – FrankH
    Oct 3, 2012 at 12:16
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    $\begingroup$ It depends on the proportion of the gap that is filled by the material. I would have to look up the how one would do the non-homogeneous calculation of the dielectric constant, which can be quite complicated for mixed materials, however the glass might have negligible effect depending on configuration. $\endgroup$
    – Freedom
    Oct 3, 2012 at 12:30

Yes, the bodies still attract. The glass may change the shape of the electric field, but it cannot simply block the electric field.

Any region of space which contains an electric charge has an electric field emanating from it.

If you create a closed surface around this charge, and calculate a surface integral of that electric field, the integral will tell you the quantity of the contained charge.

Take a look here: http://en.wikipedia.org/wiki/Gauss%27s_law

Nothing can change this. There is no "anti electric" shielding material.

Speaking of shielding, even if you place a charge into a conductive metal container, it will not block the field. It will just appear that the container itself has charges on its surface. However, if the container is grounded, then it will appear to block the field.

Bur what happens in that case is simply that the field is concentrated through the ground wire. It has not gone away!


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