双语物理asignment3.doc

Fundamentals of Physics, 2010 Assignment #3Problem 1 Chap.25/34P The plastic rod of length in the figure has a nonuniform linear charge density , where is a positive constant. (a)With at infinity, find the electric potential at on the y axis, a distance y from one end of the rod. (b) From that result find the electric field component at . (c) Why cannot the field component be found using the result of (a)? Solution(a) Divide the charged rod into many tiny segments as the differential elements of the rod. Consider one of the differential elments (see the above diagram): Position: ); Size(length): . Differential charge: Differential potential at : (Set at infinity) (according to the potential due to a point charge) The total potential at due to the rod is given by (according to the superposition principle)(b) The electric field can be obtained by the relation betweeen and the potetian distribution In this case, although we do not know the function in the whole space, we has obtained in the result (a) the y-dependence of the potential on the y-axis (namely, ). So we can find the y-component of the field on the y-axis since this component is only determined by the partitial derivative of with respect to :(c) We cannot find from the result (a) since this component is determined by the partitial derivative of with respect to . Problem 2 Chap.24/44P The figure shows a spherical shell of charge with uniform volme charge density . Plot due to the shell for distances from the center of t he shell ranging from zero to 30cm. Assume that , and .Solution According to the spherical symmetry of the charge distribution, we choose a concentric sphere of radius as the Gaussian surface. The net charge enclosed by this surface isSince the electric field at any point on the Gaussian surface must have the same magnitude and be perpendicular to the surface in any case, the electric flux through the Gaussian surface is always given by.Then the Gauss lawreduces to, so the magnitude of the electric field inside the sphere at a distance r from the centre is .Substituting , and we havewith and .From these expressions we can get the value of the field magnitude at any positions. For example, at