More about E and V Suppose we have two spherical conductors, one twice the size of the other, connected by a wire, so that charges can flow between them until the potential on each surface is the same. Suppose the left conductor has a radius R and a charge Q. 1 1 Since the potential on each must be the same, because they are connected by a wire, let’s find the charge on the right sphere, Q . 2 V on the left = V on the right kQ /1 = 1Q /(R /2) 1 Q 1 Q /(½2 Q 1 2Q 2 So the charge on the left sphere is twice the charge on the right. That makes sense. But how about the charge density, that is, charge/area Do you recall that the surface area of a sphere 2 2 2 equals 4ðR On the left sphere, we have Q /4ðR and on the r1ght, Q1/4ðR . But Q is only half 2 2 2 of Q and R is also half of R , so the charge density on the right sphere can be written like this: 1 2 1 2 2 Q /2ðR 2 = (Q /21/[4ð(R /2) 1 ] 2 = (Q /1