Physics 2 questions below. Here are some notes on how to solve the problem

2. [10] The figure below shows a long conducting cylinder of radius 6mm with a hollow core of radius 2 mm. A string runs down the center of the hollow core. There is a charge density of 1 = -2 uC/m on the string. Find the electric field at the 6 points Xia, X2a, X3a, Xib, X2b, and X3b. The points are defined in cylindrical coordinates, but give the components of E in Cartesian coordinates shown in the figure. Side View string ( insulator) with AQ air conducting cylinder 2 E = ( Ex , Ey,E2 ) r X 1Q = (0) 1 0 ) E ( xia ) ? End view a X Ia = (0 , 4 , 0 ) E ( Xia) = Xza = (0, 12, 0 ) E ( Xsa ) = -5 x * 16 = (0 , 1 , 90 " ) E ( X 15 ) = X 26 = (0 , 4 , 90' ) E ( x ,b) : X36 = (0, 12, 90" ) E ( X,b) : Guidance: There are two concepts tested by this problem. First: Gauss's Law. The flux of electric field through a surface is proportional to the charge enclosed @ = E A = Q/co. See section 19.7 of Walker or page 31 of the lecture notes. In this case consider a cylindrical surface of radius R and length L centered on the string. The field must be perpendicular to the surface. The surface area of the cylinder is A = 2 7 R L. The charge enclosed is Q = A L. Second: Charge is free to move in a conductor. If there is any net E-field, charge will redistribute until the field is E=0 inside the conductor. This can happen if charge -Q is distributed on the inner surface of the conductor and charge +Q on the outer surface. Note that in the present case Q itself is negative. See figure 19.33 of Walker for a similar example with a conducting sphere.Gauss cases TI Find Elv ) around a line of charge Sum up contributions E = EE: Ly E = SE ( b , * ) dx ugh ! Line Charge S = 2 - charge Length r T 1 radually outward A Endcaps + Cylinder 4 = 4 + 4, L A = 2TILL Endcaps Q = XL Ell surface $ = 2 = E 2 TTYL