A cylindrical capacitor has radii a and b, as in Figure 25-6. Find the radius r

Question

A cylindrical capacitor has radii a and b, as in Figure 25-6. Find the radius r such that a fraction f of the total energy is within a cylinder of radius r. Express your answer in terms of a, b, and f.

FIG. 25.6 A cross section of a long cylindrical capacitor, showing a cylindrical Gaussian surface of radius r (that encloses the positive plate) and the radial path of integration along which Eq. 25-6 is to be applied. This figure also serves to illustrate a spherical capacitor in a cross section through its center. A cylindrical capacitor has radii a and b, as in Figure 25-6. Find the radius r such that a fraction f of the total energy is within a cylinder of radius r. Express your answer in terms of a, b, and f.

Explanation / Answer

Coulomb's law that you claim is the only equation you know thus far will not help you, unless you want some painful calculus. We will be using Gauss's law Don't tell me how to approximate Pi. If you are leaving your answer in symbols, leave Pi in symbols too. If you want a numeric value, don't enter Pi's digits, use the pre-programmed value, it is more accurate. 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 r1

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