Consider a point charge, +q fixed at the origin A positive test charge,q 0 is placed at A, a distance r A Coulomb’s law determines the magnitude of repulsive force If test charge is released, it will accelerate until at point B its kinetic energy equals the electric potential energy lost

Calculus shows: U A – U B = kq 0 q/r A – kq 0 q/r B To find the change is electric potential, divide by test charge, q 0 V A -V B = kq/r A – kq/r B If the test charge is moved an infinite distance away (r B  ∞), the term kq/r B vanishes V A – V B = kq/r A We choose electric potential to be zero infinitely far from a given charge

Therefore the electric potential at an arbitrary distance: V = kq/r Recall: V represents change  from infinity to r The difference in electric potential energy: U = q 0 V U = kq 0 q/r At ∞: U = 0 Since r is a distance & positive, the potential at x = 1 m =-1 m

Therefore, V depends on the sign of the charge The potential for the positive charge increases to positive infinity near the origin and decreases to zero far away A “potential hill” Thus a positive test charge will move away from origin, as if sliding “downhill”

For negative charge  negative infinity near origin A “potential well” Again, positive charge slides downhill  toward origin Negative test charges always tend to slide “uphill” Electric potential obeys to superposition principle  the algebraic sum of the potentials due to each charge