Turns out I had made a simple mistake in my calculations. Instead of [q/(2*pi*height*epsilon_0)]*ln(a/b) I should have written [-q/(2*pi*height*epsilon_0)]*ln(a/b) because the charge of the inner cylinder is negative. Thanks for helping me realize my careless mistake.
Ok, so if I'm finding the voltage from b to a, then I need to calculate the integral of E dr from a to b? The answer of the integral is then [q/(2*pi*height*epsilon_0)]*ln(a/b). This seems right to me (unless I've made some simple mistake inbetween), yet the book says it should be ln(b/a) instead.
Hi everyone
In most problems I've seen involving capacitance of a coaxial cable, there is an inner cylinder of radius a with +Q charge and and an outer cylinder of radius b with -Q charge. The voltage is calculated from a to b. However, in this problem the inner cylinder is -Q and the outer...