10/26/2004 section 4_6 Voltage and Electric Potential empty.doc 1/2 Jim Stiles The Univ. of Kansas Dept. of EECS 4-6 Voltage and Electric Potential Reading Assignment: pp. 107-116 ()Crd⋅∫E A ()()()21Crd gr gr⋅= −∫E A, where () ()rgr=∇E Q: A: HO: Voltage and Electric Potential 1. a static electric field ()rE (a vector field). 2. an electrostatic potential field ()Vr (a scalar field!).10/26/2004 section 4_6 Voltage and Electric Potential empty.doc 2/2 Jim Stiles The Univ. of Kansas Dept. of EECS Q: A: HO: Electric Potential for Point Charge Q: A: HO: Electric Potential Function for Charge Densities Example: The Electric Dipole Q: A: HO: The Dipole Moment10/26/2004 Voltage and Electric Potential.doc 1/6 Jim Stiles The Univ. of Kansas Dept. of EECS Voltage and Electric Potential An important application of the line integral is the calculation of work. Say there is some vector field ()rFthat exerts a force on some object. Q: How much work (W) is done by this vector field if the object moves from point Pa to Pb, along contour C ?? A: We can find out by evaluating the line integral: ()rabCWd=⋅∫F A Say this object is a charged particle with charge Q, and the force is applied by a static electric field ()rE . We know the force on the charged particle is: ()()rrQ=FE Pa Pb C10/26/2004 Voltage and Electric Potential.doc 2/6 Jim Stiles The Univ. of Kansas Dept. of EECS and thus the work done by the electric field in moving a charged particle along some contour C is: ()()rrabCCWdQd=⋅=⋅∫∫FEAA A: Yes there is! Recall that a static electric field is a conservative vector field. Therefore, we can write any electric field as the gradient of a specific scalar field ()rV: ()()rrV=−∇E We can then evaluate the work integral as: ()()() ()() ()baabrrrrrrabCCWQdQV dQV VQV V=⋅=− ∇ ⋅=− −⎡⎤⎣⎦=−⎡⎤⎣⎦∫∫E AA Q: Oooh, I don’t like evaluating contour integrals; isn’t there some easier way?10/26/2004 Voltage and Electric Potential.doc 3/6 Jim Stiles The Univ. of Kansas Dept. of EECS We define: ()()ab a bVVr Vr− Therefore: ab abWQV= Q: So what the heck is Vab ? Does it mean any thing? Do we use it in engineering? A: First, consider what Wab is! The value Wab represents the work done by the electric field on charge Q when moving it from point Pa to point Pb. This is precisely the same concept as when a gravitational force field moves an object from one point to another. The work done by the gravitational field in this case is equal to the difference in potential energy (P.E.) between the object at these two points. gm=Fg..PE∆10/26/2004 Voltage and Electric Potential.doc 4/6 Jim Stiles The Univ. of Kansas Dept. of EECS The value Wab represents the same thing! It is the difference in potential energy between the charge at point Pa and at Pb. Q: Great, now we know what Wab is. But the question was, WHAT IS Vab !?! A: That’s easy! Just rearrange the above equation: ababWVQ= See? The value Vab is equal to the difference in potential energy, per coulomb of charge! * In other words Vab represents the difference in potential energy for each coulomb of charge in Q. * Another way to look at it: Vab is the difference in potential energy if the particle has a charge of 1 Coulomb (i.e., Q =1). Note that Vab can be expressed as: ()() ()rabCabVdVr Vr=⋅=−∫E A where point Pa lies at the beginning of contour C, and Pb lies at the end.10/26/2004 Voltage and Electric Potential.doc 5/6 Jim Stiles The Univ. of Kansas Dept. of EECS We refer to the scalar field ()rV as the electric potential function, or the electric potential field. We likewise refer to the scalar value Vab as the electric potential difference, or simply the potential difference between point Pa and point Pb. Note that Vab (and therefore ()rV), has units of: Joules CoulombababWVQ⎡⎤=⎢⎥⎣⎦ Joules/Coulomb is a rather awkward unit, so we will use the other name for it—VOLTS! 1 Joule 1 VoltCoulomb Q: Hey! We used volts in circuits class. Is this the same thing ? A: It is precisely the same thing ! Perhaps this will help. Say Pa and Pb are two points somewhere on a circuit. But let’s call these points something different, say point + and point - .10/26/2004 Voltage and Electric Potential.doc 6/6 Jim Stiles The Univ. of Kansas Dept. of EECS Therefore, V represents the potential difference (in volts) between point (i.e., node) + and point (node) - . Note this value can be either positive or negative. Q: But, does this mean that circuits produce electric fields? + - ()rCVd=⋅∫E AA: Absolutely! Anytime you can measure a voltage (i.e., a potential difference) between two points, an electric field must be present!10/26/2004 Electric Potential for a Point Charge.doc 1/7 Jim Stiles The Univ. of Kansas Dept. of EECS Electric Potential for Point Charge Recall that a point charge Q, located at the origin (r=0′), produces a static electric field: ()ˆ20r4rQarπε=E Now, we know that this field is the gradient of some scalar field: ()()rrV=−∇E Q: What is the electric potential function ()rV generated by a point charge Q, located at the origin? A: We find that it is: ()0r4QVrπε= Q: Where did this come from ? How do we know that this is the correct solution? A: We can show it is the correct solution by direct substitution!10/26/2004 Electric Potential for a Point Charge.doc 2/7 Jim Stiles The Univ. of Kansas Dept. of EECS ()()ˆˆ0020rr4044rrVQrQarrQarπεπεπε=−∇⎛⎞=−∇⎜⎟⎝⎠⎛⎞∂=−+⎜⎟∂⎝⎠=E The correct result! Q: What if the charge is not located at the origin ? A: Substitute r with r-r′, and we get: ()0r4r-rQVπε=′ Where, as before, the position vector r′ denotes the location of the charge Q, and the position vector rdenotes the location in space where the electric potential function is evaluated.10/26/2004 Electric Potential for a Point Charge.doc 3/7 Jim Stiles The Univ. of Kansas Dept. of EECS The scalar function ()rV for a point charge can be shown graphically as a contour plot: Or, in three dimensions as: −4−2 0 2 4x−4−2024y−4−2024x−4−2024y0.20.4volts−4−2024x−4−2024y0.20.4volts10/26/2004 Electric Potential for a Point Charge.doc 4/7 Jim Stiles The Univ. of Kansas Dept. of EECS