Physics - Electrostatics Question with Solution | TestHub
Consider first a metal sphere of radius R and total charge Q , which (because of symmetry) is distributed uniformly on its surface, i.e. the surface charge density is constant.\\ The total field effects of the charges on the small surface pieces opposite each other cancel out in pairs, so the net electric field of all the charges on the sphere is zero at any point P inside the sphere. The same is shown in figure-1. \begin{figure}[h] \begin{center}
\captionsetup{labelformat=empty} \caption{Fig. 1} \end{center} \end{figure} \begin{figure}[h] \begin{center}
\captionsetup{labelformat=empty} \caption{Fig. 2} \end{center} \end{figure} Now let us move on to the case of the electrically charged metal disc. We show that its charge distribution can be found from the uniform surface density of the charged metal sphere. Let us (notionally) 'glue into position' the charges on the surface of the sphere, and then project the points of the spherical surface perpendicularly onto a plane that contains the centre of the sphere and point P investigated earlier. As the charges cannot move, the charge , originally on the surface area , when projected onto the plane, occupies a somewhat smaller area , and so the surface charge density (with an unchanged amount of charge) increases. In addition, its distance from the point P also changes, from the original value of to the smaller one of . The corresponding thing happens with the charge originally located on the opposite side of the sphere.
A large thin metal disc of radius R is charged and carries a total electric charge of Q. The charge density on the disc at distance r from centre far away from the rim of the disc is (Neglect edge effects)
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Solution:
The electric field produced at P by the two charged pieces (projected onto the plane, and still glued to the disc) can be calculated from Coulomb's law:
In the last step, the geometrical results and were used. The same zero-field results holds for all other charge pairs, and so also for the whole charge distribution; the electric field is everywhere zero on the plane of the disc. We have obtained the interesting result that the electric field produced by all the other charges does not produce any force on the charge fixed at P, for any P. Consequently, we can even abolish the notional fixing (gluing) of the charges, because - in the absence of any force - they will not move. The question of the electrical charge distribution on the disc has thus been solved; the charge is distribute on the disc in the same way as it would be if that on a uniformly charged spherical surface were projected perpendicularly onto it. The surface charge density can be found with the help of a 'side view' sketch of a sphere of radius R , in which the particular surface pieces appear as line segments. The charges, originally situated on the area , are, after projection, at a distance r from the centre of the disc of radius R, and occupy an area at their new position. So the charge density on the disc at a distance r from its centre is
