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Prof. David R. Jackson ECE Dept. Fall 2014 Notes 4 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1.

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Presentation on theme: "Prof. David R. Jackson ECE Dept. Fall 2014 Notes 4 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1."— Presentation transcript:

1 Prof. David R. Jackson ECE Dept. Fall 2014 Notes 4 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1

2 so Electric Field The electric-field vector is the force vector for a unit charge. Units of electric field: [ V/m ] ++++++++ ++++ -------- ---- + - V 0 [ V ] E q Note: The “test” charge is small enough so that it does not disturb the charge on the capacitor and hence the field from the capacitor. 2

3 Electric Field (cont.) Note: The electric-field vector may be non-uniform. E q Point charge q in free space: 3

4 Voltage Drop The voltage drop between two points is now defined. Volta 4 C A B E (x,y,z) Comment: In statics, the line integral is independent of the shape of the path. (This will be proven later after we talk about the curl.)

5 Voltage Drop (Cont.) We now explore the physical interpretation of voltage drop. C A B E (x,y,z) q test charge F ext (x,y,z) A test charge is moved from point A to point B at a constant speed (no increase in kinetic energy). 5

6 Voltage Drop (cont.) W ext = work done by observer in moving the charge. 6 C A B E (x,y,z) q test charge F ext (x,y,z) So

7 Voltage Drop (cont.) so But we also have or 7

8 Physical Interpretation of Voltage 8 The voltage drop is equal to the change in potential energy of a unit test charge. Point A is at a higher voltage, and hence a higher potential energy, than point B. ++++++++ ++++ -------- ---- + - V 0 [ V ] E q = 1 A B

9 Comments 9  The electric field vector points from positive charges to negative charges.  Positive charges are at a higher voltage, and negative charges are at a lower voltage. ++++++++ ++++ -------- ---- + - V 0 [ V ] E

10 Review of Doing Line Integrals 10 In rectangular coordinates, Hence Note: The limits are vector points. so Each integrand must be parameterized in terms of the respective integration variable. This requires knowledge of the path C.

11 Example Find: E Assume: ++++++++ ++++ -------- ---- + - V 0 [ V ] E x = 0 x = hx = h A B 11 Ideal parallel-plate capacitor assumption.

12 Example (cont.) Evaluate in rectangular coordinates: 12

13 Example (cont.) Hence, we have Recall that 13

14 Example A proton is released at point A on the top plate with zero velocity. Find the velocity v ( x ) of the proton at distance x from the top plate (at point B). 14 Conservation of energy: h = 0.1 [ m ] V 0 = 9 [ V ] ++++++++ ++++ -------- ---- + - V 0 [ V ] E x = 0 x = hx = h Proton A B x = x x = 0

15 Example (cont.) Hence: 15

16 Example (cont.) Hence: h = 0.1 [ m ] V 0 = 9 [ V ] q = 1.602 x 10 -19 [C] m = 1.673 x 10 -27 [kg] (See Appendix B of the Hayt & Buck book or Appendix D of the Shen & Kong book.) 16

17 Reference Point  A reference point R is a point where the voltage is assigned.  This makes the voltage unique at all points.  The voltage at a given point is often called the potential . R C r = (x,y,z) Note:  0 is often chosen to be zero, but this is not necessary. 17 Integrating the electric field along C will determine the potential at point r.

18 Example Find the potential on the left terminal (cathode) assuming R is on right terminal (anode) and  = 0 at the reference point. 18 9 [ V ] A B + - A B

19 Find the potential function at x, assuming that the reference point R is on the bottom plate, and the voltage at R is zero.Example 19 ++++++++ ++++ -------- ---- + - V 0 [ V ] E x = 0 x = hx = h r (x, y, z) Also, Hence


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