Electric Potential Electric Potential Energy Work done by Coulomb force when q 1 moves from a to b: b a FEFE r dr ds q 2 (-) q 1 (+) rara rbrb

The important point is that the work depends only on the initial and final positions of q 1. In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force. b a FEFE r dr ds q 2 (-) q 1 (+) rara r ab Electric Potential Energy

E A charged particle in an electric field has electric potential energy. It “feels” a force (as given by Coulomb’s law). F Electric Potential Energy It gains kinetic energy and loses potential energy if released. The Coulomb force does positive work, and mechanical energy is conserved.

Electric Potential Dividing W by Q gives the potential energy per unit charge. V AB, is known as the potential difference between points A and B. The electric potential V is independent of the test charge q 0.

If V AB is negative, there is a loss in potential energy in moving Q from A to B; the work is being done by the field. if it is positive, there is a gain in potential energy; an external agent performs the work Electric Potential E F

Electric Potential V AB is the potential at B with reference to A V B and V A are the potentials (or absolute potentials) at B and A

If we choose infinity as reference the potential at infinity is zero;the electric potential of a point charge q is Electric Potential The potential at any point is the potential difference between that point and a chosen point in which the potential is zero.

Things to remember about electric potential:  Electric potential difference is the work per unit of charge that must be done to move a charge from one point to another without changing its kinetic energy. ●Sometimes it is convenient to define V to be zero at the earth (ground).  The terms “electric potential” and “potential” are used interchangeably.  The units of potential are joules/coulomb:

Example: a 1  C point charge is located at the origin and a -4  C point charge 4 meters along the +x axis. Calculate the electric potential at a point P, 3 meters along the +y axis. q2q2 q1q1 3 m P 4 m x y Thanks to Dr. Waddill for the use of these examples.

Example: how much work is required to bring a +3  C point charge from infinity to point P? q2q2 q1q1 3 m P 4 m x y q3q3 0 The work done by the external force was negative, so the work done by the electric field was positive. The electric field “pulled” q 3 in (keep in mind q 2 is 4 times as big as q 1 ). Positive work would have to be done by an external force to remove q 3 from P.

Electric Potential of a Charge Distribution Collection of charges: Charge distribution: P is the point at which V is to be calculated, and r i is the distance of the i th charge from P. P r dq Potential at point P.

Electric Potential of a Charge Distribution

Example: A rod of length L located along the x-axis has a total charge Q uniformly distributed along the rod. Find the electric potential at a point P along the y-axis a distance d from the origin. Thanks to Dr. Waddill for this fine example. y x P d L dq x dx r =Q/L dq= dx

y x P d L dq x dx r A good set of math tables will have the integral:

Example: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring. P R dQ r x x Every dQ of charge on the ring is the same distance from the point P.

P R dQ r x x

Example: A disc of radius R has a uniform charge per unit area  and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center. P r dQ x x R The disc is made of concentric rings. The area of a ring at a radius r is 2  rdr, and the charge on each ring is  (2  rdr). We *can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.

P r dQ x x R

P r x x R Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 3?

MAXWELL'S EQUATION The line integral of E along a closed path is zero This implies that no net work is done in moving a charge along a closed path in an electrostatic field

MAXWELL'S EQUATION Applying Stokes's theorem Thus an electrostatic field is a conservative field

Electric Potential vs. Electric Field Since we have As a result; the electric field intensity is the gradient of V The negative sign shows that the direction of E is opposite to the direction in which V increases

If the potential field V is known, the E can be found Electric Potential vs. Electric Field

Example: In a region of space, the electric potential is V(x,y,z) = Axy 2 + Bx 2 + Cx, where A = 50 V/m 3, B = 100 V/m 2, and C = -400 V/m are constants. Find the electric field at the origin