Notes 4 Electric Field and Voltage

Transcription

1 ECE 3318 pplied Electricity and Magnetism Spring 2018 Prof. David R. Jackson Dept. of ECE Notes 4 Electric Field and Voltage Notes prepared by the EM Group University of Houston 1

2 Electric Field V 0 [ V] - q E so F = qe E 1 q F 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. The electric-field vector is the force vector for a unit charge. Units of electric field: [V/m] 2

3 Electric Field (cont.) Note: The electric-field vector may be non-uniform. Point charge q in free space: q E E q = rˆ 2 4πε 0r 3

4 Voltage Drop The voltage drop between two points is now defined. C Volta E( xyz,, ) V V V E dr ( ) ( ) 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.) 4

5 Voltage Drop Here all of the mathematical elements are defined. r dr C r z E( xyz,, ) x y V E dr 5

6 Voltage Drop (Cont.) We now explore the physical interpretation of voltage drop. F ext q ( xyz,, ) ( test charge) C E( xyz,, ) test charge is moved from point to point at a constant speed (no increase in kinetic energy). F E = qe ext E F = F = qe 6

7 Voltage Drop (cont.) F ext q ( xyz,, ) ( test charge) C E( xyz,, ) Define: W ext = work done by observer in moving the charge. ext ext W = F dr = q E dr = q E dr Hence ext W = qv 7

8 Voltage Drop (cont.) From the last slide: ext W = qv ut we also have ext W = PE PE ( ) ( ) so qv = PE PE ( ) ( ) or 1 V = PE PE q ( ) ( ) 8

9 Physical Interpretation of Voltage Result: V = PE PE q = 1 ( ) ( ) ( ) Conclusion: The voltage drop is equal to the change in potential energy of a unit test charge. The voltage at a point physically means the potential energy of a unit test charge at that point. 9

10 Physical Interpretation of Voltage Example: Point is at a higher voltage, and hence a higher potential energy, than point. V 0 [ V] - q = E Parallel-plate capacitor V = PE PE q = 1 ( ) ( ) ( ) 10

11 Comments 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. The electric field points from higher voltage to lower voltage. V 0 [ V] - E

12 Review of Doing Line Integrals In rectangular coordinates, V = E dr E= xe ˆ ye ˆ ze ˆ x y z C r = xˆx yˆ y zˆz dr = xˆdx yˆdy zˆdz Hence so ( ) V = E dx E dy E dz x y z x y z (,, ) (,, ) (,, ) V = E x y z dx E x y z dy E x y z dz x y z x y z Note: The limits are vector points. Each integrand must be parameterized in terms of the respective integration variable. This requires knowledge of the path C. 12

13 Example V 0 [ V] E x = 0 x = h Find: E E x, y, z = xe ˆ ssume: ( ) 0 Ideal parallel-plate capacitor assumption V ( ) = E x, y, z dr = V0 [ V] 13

14 Example (cont.) ( ) 0 V V = E x, y, z dr = V [ ] Evaluate in rectangular coordinates: E= xe ˆ ye ˆ ze ˆ x y z dr = xˆdx yˆdy zˆ dz E dr = Ex dx Ey dy Ez dz x = x = x V E dx V [ ] 0 V 14

15 Example (cont.) x = x = x V E dx V h 0 E dx 0 = V 0 0 V 0 [ V] E (,, ) = xe ˆ 0 E x y z x = 0 x = h E h = V E = V / h Recall that E x, y, z = xe ˆ ( ) 0 Hence, we have E xyz V = ˆ h,, x [ V/m] ( ) 0 15

16 Example V 0 [ V] V 0 = 9V [ ] - h = 0.1 x = 0 [ m ] Proton E x = x x = 0 x = h proton is released at point on the top plate with zero velocity. Find the velocity v(x) of the proton at distance x from the top plate (at point ). Conservation of energy: ( ) ( 0) = ( 0) ( ) KE x KE PE PE x 1 2 ( ) = ( 0) ( ) = ( 0) ( ) 2 mv x PE PE x q V V x 16

17 Example (cont.) From last slide: 1 ( ) = ( 0) ( ) 2 2 mv x q V V x x x V0 V0 V ( 0) V ( x) = E dr = E dx = dx = x h h 1 2 Hence: ( ) x 0 0 V = h 2 0 mv x q x so 2qV = hm ( ) 0 v x x V 0 [ V] V 0 = 9V [ ] x = 0 h = 0.1 [ m] Proton - E x = x x = 0 x = h 17

18 Example (cont.) 2qV = x hm ( ) 0 v x V 0 = 9 [V] h = 0.1 [m] q = x [C] m = x [kg] (See ppendix of the Hayt & uck book or ppendix D of the Shen & Kong book.) Hence: 6 ( ) = v x x [ m/s] 18

19 Reference Point 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 Φ. r = ( xyz,, ) R Φ ( R) =Φ0 C Integrating the electric field along C will determine the potential at point r. Note: Φ 0 is often chosen to be zero, but this is not necessary. 19

20 - Example 9 [ V] V = 9 Find the potential on the left terminal (cathode) assuming R is on the right terminal (anode) and Φ = 0 at the reference point. 9 [ V] - ( ) ( ) 9 V =Φ Φ = 9 [ V] Φ ( ) = V 9[ ] Φ ( R) 0 20

21 - 9 [ V] V = Example 9 Find the potential on the right terminal (anode) assuming R is on the left terminal (cathode) and Φ = 0 at the reference point. - 9 [ V] 9 [ V] ( ) ( ) 9 V =Φ Φ = ( ) V Φ = 9[ ] Φ ( R) 0 This is the more usual choice for the reference point (on the negative terminal). 21

22 Example V 0 [ V] - C r( xyz,, ) E x = 0 x = h Φ ( R) 0 Find the potential function at x, 0 < x < h, assuming that the reference point R is on the bottom plate, and the voltage at R is zero. x x V0 V0 V =Φ( 0) Φ ( x) = E dx = dx = x h h x 0 0 Hence lso, ( 0) ( ) ( 0) Φ Φ h = V Φ = V V0 Φ ( x) = V x h 0 V 0 0 [ ] 22