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I would like to ask a conceptual question. Consider an uncharged conductor suspended in mid air, is there an electric field that I can introduce to move the conductor,i.e cause a net force on the conductor. I already know that the electrons will move around to the net E-field inside the conductor zero. But would there be a net force, or will the forces on the individual charged particle cancel out. Please give an example if a net force is caused.

alexan123
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3 Answers3

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If the conductor is not fixed in space, The external electric field E will induce charges on the surface of the sphere. The charges negative charges will be attracted and the positive charges repelled. Thus, the conductor will move towards the source of the E field. Cancellation of the external field occur inside the 'metal' of the conductor. The surface charges do feel a force, which will cause the sphere to move as described.

Lelouch
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  • There won't be a net force on the sphere or any shaped uncharged conductor. – Jan Bos Sep 11 '16 at 06:05
  • WHy not? The induced charges will be attracted. – Lelouch Sep 11 '16 at 06:21
  • I stand corrected. There could be a net force if the external field is non uniform in the sphere. – Jan Bos Sep 11 '16 at 07:46
  • I dont understand your claim. The electric field will cause separation of positive and negative centres, irrespective of the uniformity of the field. – Lelouch Sep 11 '16 at 08:09
  • Please see http://physics.stackexchange.com/questions/90556/does-a-conductor-of-total-charge-zero-placed-in-a-uniform-external-electric-fiel?rq=1 – Jan Bos Sep 11 '16 at 08:20
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Yup. Consider for example a conducting sphere of radius $R$ with its center at the origin, and a charge $q$ placed on the x-axis at $x=d>R$. It's a well known result that after the electrostatic condition has been reached, we can use the method of image charges and conclude that the final configuration of the fields outside the sphere are equivalent to those of a pair of charges: one $q'=-\frac{R}{d}q$ placed at $x=d'=\frac{R^2}{d}$ and another $-q'$ placed at the origin. From that we can conclude that the electric field at $x=d$ is $$\overrightarrow{E}=\left(-\frac{kq'}{d^2}+\frac{kq'}{(d-d')^2}\right)\hat{x}=kq\left(\frac{R}{d^3}-\frac{Rd}{(d^2-R^2)^2}\right)\hat{x},$$ and therefore the charge feels a force $$\overrightarrow{F}=q\overrightarrow{E}=kq^2\left(\frac{R}{d^3}-\frac{Rd}{(d^2-R^2)^2}\right)\hat{x}.$$ But we know from Newtons Third Law that there must be a reaction force $-\overrightarrow{F}$ acting on the conducting sphere, which is exactly what was to be shown.

Gabriel Golfetti
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An electrical conductor creates a region in space that is devoid of internal electric fields. An electric field contains stored energy, with the density of that energy proportional to the square of the field. So, the electric field loses stored energy (does mechanical work) on any conductor that moves from a low-E-field region to a high-E-field region.

Conductive particles are attracted to the highest field strength region, and the force on the particles is proportional to the gradient of the square of the field absolute value. Nonconductive particles which have a positive dielectric constant are also weakly attracted to high electric fields.

As for an example, one need only rub a plastic rod on wool, and see that it attracts a bit of aluminum foil...

Whit3rd
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