Tanya Sharma said:
This is the point I am unable to comprehend...I understand how the shell is shielding but why does the shell shields the central charge ?
According to Gauss Law , charge q will induce charge –q on the inner surface of the shell ,so that charge on the outer surface of the shell is Q+q .
This is correct and is found with Gaussian surface in the interior of the conducting shell.
Central charge q will be under the influence of electric field due to the charge –q on the inner surface of the shell .By symmetry ,the force should be zero .
However you cannot draw this conclusion by argument of symmetry. The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell and the charge outside the shell.
When the outer charge is not present, you *can* draw the symmetry conclusion. This is because the charge on the outer and inner surfaces of the shell will distribute evenly. This must happen, otherwise there is a tangential field on the surfaces of the shell that will cause surface charge to continue to move. So in the static case, the surface charges must evenly distribute. Then you can look at the interior of the shell and state the field must be directed toward the center of the shell by symmetry because of the even distribution of charge on the shell surfaces (there is no preferential location inside the shell as a function of theta). Only then can you use Gauss's Law to find the interior field strength. Gauss's Law cannot be fruitfully applied until you can make this symmetry argument. It is important to note this: all charges on the outer surface of the shell and the inner surface of the shell are acting on the point charge in the interior, however their distribution causes the field in the interior to be directed radially toward the shell center. This allows Gauss's Law to be applied fruitfully to find the field strength. I think I am rambling here, but it is to get to this point that many students misunderstand: Gauss's Law does not state that the field on its surface is only generated by charge interior to its surface -- this is absolutely incorrect. The field at any point in space is determined by the distribution of charges in all space. In this example, those charges outside the Gaussian surface affect the field inside the shell by controlling the field direction, making it radial.
In the current example with a charge also placed outside the shell, charge on the outer shell will not be distributed evenly -- a higher concentration of charge on the outer shell will appear nearest the outside charge and, due to conservation of charge, a lower concentration of charge will appear on the side of the outer shell diametrically opposite.
The symmetry argument is gone.
Instead, as haruspex points out, you can take advantage of the fact the shell is a perfect conductor and therefore sits at the same potential throughout. So in the interior of the shell we can rely on Laplace's Equation ∇
2V = p
v / ε. This equation says the field on the inside of the shell depends only on the charge distribution inside the shell and the voltage on the boundary of the inner shell. Since the voltage is everywhere the same on the inner shell, the field must be radially directed to the center and again Gauss's Law can be used to find the field inside the shell. And you find the field is independent of any outside charge.
Since you know the field inside the shell is the same whether or not the outside charge exists, you know that every charge in space exerts a force on the charge inside the shell, you know exactly the force the outside point charge is exerting on the inside point charge (coulomb's law), what kind of force must the charge distribution on the outer shell be exerting on the inner point charge in order for it to experience a net force of zero?
