Two charges are held fixed in an x-y plane. Charge Q=-1*10^-9 C is located at the origin, q=3*10^-9 C is located at (0,2). Determine the voltage (v_ba) between point 'b' located at (3,0) and point 'a' located at (1,0). Assume all distances on the plane are given in meters. The answer is given as 1.41 V but I can't seem to get that I try to use the integrated formula: v= 9x10^9Qq[(1/R_b) - (1/R_a)] but I'm way off And if you could explain to me what higher potential means, and does the higher potential need to be negative or positive? or can it change depending on the current? That might not have made sense, but this book is awful and I'm pretty lost, any help would be greatly appreciated!