Electric Potential Energy Work done by Coulomb force

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

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

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

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 q 0.

Electric Potential +++++++ If VAB 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 + E F ----------

Electric Potential VAB is the potential at B with reference to A VB and VA are 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 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. y P 3 m q 1 4 m q 2 x

Example: how much work is required to bring a +3 C point charge from infinity to point P? 0 y q 3 P 3 m q 1 4 m q 2 x 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: 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 P Potential at point P. r

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 d dq= dx r dq dx x L x

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

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

d. Q 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. d. Q r P R x x 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.

d. Q r P R x x

d. Q r P R x x 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

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) = 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