Work and Voltage We studied electric force and electric field…now we can expand our discussion to include WORK and POTENTIAL ENERGY just as we did with the gravitational field: Recall:W grav = Fd = (mg)d Let d =  h, the change in height; then, W grav = -mg  h hihi hfhf Why the minus sign?

Work and Voltage We insert the negative sign because we want the WORK due to gravity to be POSITIVE when the object falls. hihi hfhf W grav = -mg  h W grav = -mg  h f -h i ) W grav = -mgh f + mgh i Since h f is smaller than h i, our answer is now positive

Work and Voltage For the exact same reason (h f is less than h i ), the object loses PE as it falls; this we expect. Note that:  PE = mg  h = mg(h f – h i ) = mgh f - mgh i This will be a negative answer hihi hfhf

Work and Voltage Therefore we found W = -  PE We can apply the same analysis to the electric field…with one condition: The following analysis applies only to POSITIVE charges …

Work and Voltage There are TWO cases to be analyzed 1.Spherical charge distribution Fairly specialized + The van de Graaf generator is one example.

Work and Voltage Uniform charge distribution Practical application: the field inside electrical wires A wire may seem small to you, but if the wire had a diameter that equaled the distance from the Earth to the Sun (93,000,000 mi), an electron would be smaller than the period at the end of this sentence. Exposed end of wire metal plate

Work and Voltage 1.Spherical charge distribution 2. Uniform charge distribution

Work and Voltage 1.Spherical charge distribution 2. Uniform charge distribution Let d = r

Work and Voltage 1.Spherical charge distribution 2. Uniform charge distribution Let d = r Now we can substitute in an expression for FORCE (F)

Work and Voltage 1.Spherical charge distribution 2. Uniform charge distribution Let d = r What are two formulas for “F”?

Work and Voltage 1.Spherical charge distribution 2. Uniform charge distribution Let d = r

Work and Voltage Electric Field Force Work F = |q|E W = FdW = qEd Equations in red are for uniform electric fields Equations in black are for spherical charges

Work and Voltage Given a uniform Electric Field: W elec = Fd A B for uniform electric fields only 2. Uniform charge distribution

Work and Voltage Given a uniform Electric Field: A B W elec = qEd for uniform electric fields only 2. Uniform charge distribution

Work and Voltage Distance is measured from the charge source, so B > A (unlike gravitation) 2. Uniform charge distribution A B

Work and Voltage Distance is measured from the charge source, so B > A (unlike gravitation) If we let the charge “fall” ( Electric Field does work ) it will move from A to B and we get + work done: 2. Uniform charge distribution A B W elec = qE(d B – d A )

Work and Voltage A B If we push the charge from B to A (recall B > A), we get negative work…meaning that energy is stored. W elec = qE(d A -d B ) final initial 2. Uniform charge distribution = negative

Work and Voltage  Work stored in Electric Field is negative.  Work done by the Electric Field is positive. W elec = qE(d B – d A ) W elec = qE(d A -d B )

Work and Voltage Work F = qE W = Fd W = qEd Equations in red are for uniform electric fields Equations in black are for spherical charges

Work and Voltage Since we previously determined that W = -  PE We can write: -  PE = qEd  PE = -qEd When work is done, PE is lost

Work and Voltage Now we define voltage: Voltage is the Energy available to each coulomb of charge (it is NOT the total energy – that’s “W”)

Work and Voltage Why isn’t there a ‘gravitational voltage’? What the heck does this mean?

Work and Voltage Example: For Earth, g = 9.8 N/kg. What is the gravitational voltage at a height of 3.0 meters?

Work and Voltage Example: For Earth, g = 9.8 N/kg. What is the gravitational voltage at a height of 3.0 meters?  V grav = gh = (9.8 N/kg)(3.0 m) = 29.4 Joules/kg Meaning: every kg of mass will provide 29.4 Joules of energy at this height

Work and Voltage For a spherical charge: For a uniform Electric field:

Work and Voltage F = qE W = Fd W = work (J); V = voltage(Volts); E = Electric Field (N/C); F = force (N); q = charge (Coul); d = distance(m) W = qEd V = -Ed

Work and Voltage So now we examine VOLTAGE for a uniform Electric Field: High Voltage Low Voltage The negative sign means the voltage gets more negative (decreases) as you move away from the positive charges.

Work and Voltage Example: Find the voltage that the + charge moves through given an Electric Field = 1000 N/C and a distance, d = 0.10 m A B 0.1 m  V = -Ed = -(1000 N/C)(0.1 m) = -100 Nm/C = -100 Joules/C = -100 volts This tells us that every Coulomb of charge will lose 100 Joules of energy. One-tenth of a Coulomb will lose 10 Joules…etc.

Work and Voltage We can relate Work and Voltage in the following way: Since  V = -Ed, we can write

Work and Voltage F = qE W = Fd W = work (J); V = voltage(Volts); E = Electric Field (N/C); F = force (N); q = charge (Coul); d = distance(m) W = qEd W = -qV V = -Ed

Work and Voltage Example 2: A proton “falls” from 12 volts to 0 volts. What work was done (or what energy was used)? A B High Voltage (12 V) Low Voltage (0 V) W = -q(  V) W = -q(V f – V i ) W = -(1.6x C)( V) W = x Joules

Work and Voltage Example 3: An electron (N.B.!) moves from 0 to 12 volts. What energy was used (or what work was done)? A B High Voltage (12 V) Low Voltage (0 V) W = -q(  V) W = -q(V f – V i ) W = -(-1.6x C)( V) W = x Joules The electron falls “up”!

Work and Voltage Other names for Voltage: Potential difference (NOT potential energy difference) Electric potential

Work and Voltage In an electric field, the voltage at the equal distance from the source is always the same: the blue lines are called “lines of equipotential”. Voltage 1 Voltage 2 Voltage 3

Work and Voltage + Voltage 1 Voltage 2 For spherical sources, the “lines of equipotential” are also spherical.