ECE 1100: Introduction to Electrical and Computer Engineering Notes 17 Electric Field, Voltage, and Power Spring 2008 David R. Jackson Professor, ECE Dept.

Electric Field + - V0V x A B h q (“test charge”) FxFx F x = force on test charge (proportional to q ) The electric field strength is the force per unit charge

Electric Field (cont.) For a “unit” test charge ( q = 1 [C]): We can also write this as

Electric Field (cont.) The electric field vector: The direction of the vector is the direction of the force on a q = 1 [ C ] test charge. The magnitude of the vector is the magnitude of the force on a q = 1 [ C ] test charge. In general, For a “unit” test charge ( q = 1 [ C ]) :

Electric Field (cont.) Inside a parallel-plate capacitor, the electric field vector is (approximately) a constant. Parallel-plate capacitor + - V0V x A B h E

Electric Field (cont.) The electric field vector is pointed in the radial direction in spherical coordinates, and decreases with distance from the point charge. q Point charge

Voltage Drop V AB  voltage (at A ) – voltage (at B ) V AB  E x h Note: the unit of electric field is volts/meter + - V0V0 x A B h

Voltage Drop V AB  E x h Notes:  The electric field vector points from the positive charges to the negative charges.  The electric field vector points from the higher voltage to the lower voltage. + - V0V0 x A B h

Voltage Symbol (Reference Direction) V  voltage (at + sign) – voltage (at – sign) Voltage drop symbol (reference direction): +- V Note: V may be numerically positive or negative.

Given: V = -3 [ V ] Example Find: V AB +- V BA V BA = V(B) – V(A) = V = -3 [V] V AB = -V BA = - (-3) V AB = 3 [V]

Voltage Source: Battery V AB = V ( A ) – V ( B ) = V 0 Battery (constant voltage) V 0 = battery voltage A B V0 [V]V0 [V] anode (+ terminal) cathode (- terminal) Note: V 0 is always positive for a battery.

Voltage Source Time-varying voltage source v AB = v A ( t ) – v B ( t ) = v s ( t ) + - v s ( t ) A B Note: A is not always at the higher voltage! (i.e., v s ( t ) might be negative for some value of time).

Example (Find the Voltage) V = V AB = +9 [V] A 9 [V] B + - V V = V AB = -V BA = -9 [V] B 9 [V] A + - V + - 3sin(t) [V] + - v v = 3sin(t) [V] + - 3sin(t) [V] - + v v = -3sin(t) [V]

Example (Find V ) V = V AB = - V BA = -9 [V] 9 [V] 9 9  + - V A B Note: there is no loss of voltage across an ideal wire (zero resistance).

Relation Between Voltage and Work + - V0V0 x A B h q (“test charge”) FxFx The test charge is moved from A to B :  W = energy (work) given to particle by the electric field

Relation Between Voltage and Work (cont.) The voltage drop between two points is the energy given to a unit test charge ( q = 1 [ C ]) that moves between the two points. or

Example An electron moves from the negative to positive terminal of a 12 [ V ] battery (starting from rest). How much speed does it pick up? - + A B or q =  [C] V AB = - 12 [V] m =   [kG] v =  10 6 [m/s]

Power Dissipation device i + - v Assume a “Passive Sign Convention” : The reference direction for current points through the device from + to -. Let P abs = power absorbed by device

Power Dissipation (cont.) In time  t : device i + - v qq A B Note: For a resistor, the energy given to the charges eventually ends up as heat. Let  W = energy given to charge  q as it moves from A to B.

Power Dissipation (cont.) Note: the power delivered (provided) by a device is the negative of the power absorbed by the device.

Example Find P abs by resistor 1 [A] V 9 9  + - V

Example (done a different way) Find P abs by resistor V 9 9  -1 [A] + - V = -9 [V]

Example Find P abs by battery or V 9  -1 [A]

Example Find P abs by each element 3 [A] [V] 3 [A] [V]

Example (cont.) Find P abs by each element 2 [A] [V] - 2 [A] [V]

Example Find P abs ( t ) and the energy  W abs, defined as the energy that is absorbed by the device for 3 < t < 4 [s]. + - v x ( t ) i x ( t ) Given:

Example (cont.) + - v x ( t ) i x ( t ) The energy absorbed in the time interval (3,4) is:

Example (cont.) + - v x ( t ) i x ( t )