T he figure shows a cylindrical capacitor, it consists of a solid metal rod of r
ID: 2219888 • Letter: T
Question
The figure shows a cylindrical capacitor, it consists of a solid metal rod of radius r1 surrounded by a metal cylinderwith inner radius r2 and outer radius r3. suppose the capacitor has a length L (with L very large). Also suppose a charge +Q is placed on the inner rod and a charge -Q is placed on the outer cylinder.
a) find the electric field inside the rod (i.e. for r < r1). Use only r, Q, L, r1, r2, r3 and ep (ep forepsilon). For pi use 3.142.
b) Find the electric field in the space between the rod and the cylinder (r2 > r > r1). Use only r, Q, L, r1, r2, r3 and ep ( ep for epsilon).
c)Find the electric field outside the outer cylinder (r > r3)
Explanation / Answer
Part A: The electric field inside the rod. Because the rod is a metal, all the charge is concentrated at the outer surface. For this reason, if you make your Gaussian surface inside the rod, you will have no charge enclosed, and therefore zero electric field within the rod. ---------------------- Part B: The electric field in the spacing. This part is more interesting. As per Gauss's Law, the total electric flux over a cylindrical surface enclosing the rod is PHI_E = Q/epsilon0 Since it is positive charge, PHI_E is positive. The cylindrical surface of radius R encloses the rod. We neglect the top and bottom surfaces, because the electric field doesn't point in the axial direction (or at leas none which is of significant concern). The definition of PHI_E PHI_E = int(E·dA) For the special case of the E-field being uniform over the Gaussian surface, and perpendicular with the surface normal vector, the integration and dot product vector calculus reduce to the following algebra: PHI_E = E*A The surface area of a cylinder side: A = 2*Pi*r*L Re-construct: E*2*Pi*r*L = Q/epsilon0 Solve for E, and we get our concluding formula: E = Q/(2*Pi*epsilon0*r*L) For r1Related Questions
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