THE ELECTRIC POTENTIAL OF POINT CHARGES

Consider a point charge,

+q fixed at the originA positive test charge,q0

is placed at A, a distance rA

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:

UA – UB = kq0q/rA – kq0q/rB

To find the change is electric potential, divide by test charge, q0

VA-VB = kq/rA – kq/rB

If the test charge is moved an infinite distance away (rB ∞), the term kq/rB vanishesVA – VB = kq/rA

We choose electric potential to be zero infinitely far from a given charge

Therefore the electric

potential at an arbitrary distance:V = kq/rRecall: V represents change

from infinity to rThe difference in electric

potential energy: U = q0VU = kq0q/rAt ∞: U = 0Since r is a distance &

positive, the potential at x = 1 m =-1 m

Therefore, V depends on the sign of the chargeThe 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