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A spherical conducting shell of inner radius r_(1) and outer radius r_(2) has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain. |
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Answer» Solution :(a) `-q(4PI r_(1)^(2)), (Q+q)//(4pi r_(2)^(2))` (b) By Gauss’s law, the NET charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric FIELD inside must be zero. The cavity may have POSITIVE and negative charges with total charge zero. To dispose of this possibility, take a closed loop, PART of which is inside the cavityalong a field line and the rest inside the conductor. Since field inside the conductor is zero, this gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence, there are no fieldlines inside the cavity (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape. |
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