Electric Potential Electric Potential Energy

Electric Potential Electric Potential Energy Electric Potential Energy Work done by Coulomb force when q1 moves from a to b: r b dr q1(+) ds FE rb q1(+) a ra q2 (-)

Electric Potential Energy r b dr q1 (+) ds FE rab a The important point is that the work depends only on the initial and final positions of q1. ra q2 (-) In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force.

+ Electric Potential Energy A charged particle in an electric field has electric potential energy. It “feels” a force (as given by Coulomb’s law). + + + + + + + + + + + + + + F E 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. VAB, is known as the potential difference between points A and B. The electric potential V is independent of the test charge q0.

+ Electric Potential + + + + + + + + + + + + + + If VABis 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 F E - - - - - - - - - - - - - - - - - - -

Electric Potential VABis the potential at B with reference to A VBand VAare the potentials (or absolute potentials) at B and A

Electric Potential If we choose infinity as reference the potential at infinity is zero;the electric potential of a point charge q is The potential at any point is the potential differencebetween 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. y P 3 m x q2 q1 4 m 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? 0 y q3 P 3 m x q2 4 m q1 The work done by the external force was negative, so the work done by the electric field was positive. The electric field “pulled” q3 in (keep in mind q2 is 4 times as big as q1). Positive work would have to be done by an external force to remove q3 from P.

Electric Potential of a Charge Distribution Collection of charges: P is the point at which V is to be calculated, and ri is the distance of the ith charge from P. Charge distribution: dq r P 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. y =Q/L P r dq=dx d dq x dx x L Thanks to Dr. Waddill for this fine example.

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

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. dQ Every dQ of charge on the ring is the same distance from the point P. r R P x x

dQ r R P 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. dQ 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). r P x x R We *can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.

dQ r P 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? dQ r P x x R

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 closedpath 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 directionof E is opposite to the direction in which V increases

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

Example: In a region of space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are constants. Find the electric field at the origin

Electric Potential Electric Potential Energy