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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 is +Q. In this case, would I still find the voltage from a to b, or would I find the voltage from b to a? The answer key suggests that the voltage is calculated from a to b, but I'm not sure why.