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Electric Potential Electric Potential Energy Electric Potential Energy Work done by Coulomb force when q1 moves from a to b: r dr q1 (+) ds FE rb a q1 (+) ra q2 (-) b Electric Potential Energy r dr The important point is that the work depends only on the initial and final positions of q1. a ra FE b ds rab q2 (-) In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force. q1 (+) 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 q0. 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 1 q V r . 40 r 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: 1 joule 1 volt = 1 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 q1 q 2 qi VP = k = k + i ri r1 r2 P 3m q1 4m q2 Thanks to Dr. Waddill for the use of these examples. x -6 -6 1×10 -4×10 9 = 9×10 + 3 5 = - 4.2×103 V Example: how much work is required to bring a +3 C point charge from infinity to point P? 0 y Wexternal q3 VP V q3 P Wexternal 3 106 4.2 103 3m q1 4m q2 x Wexternal 1.26 103 J 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 1 Collection of charges: VP 40 qi i r . i P is the point at which V is to be calculated, and ri is the distance of the ith charge from P. Charge distribution: 1 dq V . 40 r Potential at point P. dq 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 dq dx dV k k r x2 d2 L V dV 0 Thanks to Dr. Waddill for this fine example. V L 0 dx Q L dx k k 2 2 L 0 x2 d2 x d y A good set of math tables will have the integral: P d r dq dx x L x dx x d 2 2 ln x x 2 d 2 kQ L L2 d 2 V ln L d 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 r R P x Every dQ of charge on the ring is the same distance from the point P. x dq dq dV k k r x2 R2 V ring dV k ring dq x2 R2 dQ r R P x x V V k x R 2 2 kQ x2 R2 ring dq 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 r P R x x The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr). We *can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R. dVring k2rdr x2 r2 dQ r P R x x 1 V dV ring 40 V x2 r2 20 R 0 20 2rdr ring x 2 r 2 20 Q x R x 20 R 2 2 2 Q R 2 R 0 rdr x2 r2 x2 R2 x dQ r P R x x Q V 20 R 2 x2 R 2 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) = 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 V E x (0, 0, 0) Ay 2 2Bx C C (0,0,0) x (0,0,0) V E y (0, 0, 0) (2Axy) (0,0,0) 0 y (0,0,0) V E z (0, 0, 0) 0 z (0,0,0) V E(0,0,0) 400 ˆi m