Work in an electric field:
Recall that the work required to move a charge across an electric field is given by W = q∆V where q is the amount of charge moved and ∆V is the difference between the electric potential at one end of the electric field and the other.
1. How much work is required to move +12 C of charge from the +5 V plate of a parallel plate capacitor to the -5 V plate?
*W = q∆V = 12 C • 10 V = 120 J (note that a V is a J/C so if you watch units, C • J/C = J)(additional note, a volt is a joule per coulomb because potential is the amount of work that must be done to move one unit of charge from the ground to that potential)
We don't usually think of this kind of charge transfer in a capacitor. This is much more likely to occur in a simple direct current circuit where a certain amount of charge moves out of the positive terminal of the battery and, to avoid accumulating charge, into the negative end of the batter. Remember that current is defined as the rate of flow of charge, that is: I = ∆q/t. Solved for charge we have ∆q = I•t.
2. A 175 mA current operates for 2.5 minutes (t must be in seconds) in a 12 Ω resistance.
a. how much charge is transferred?
I = ∆q/t so ∆q = I•t = .175 A • 2.5 min • (60 s/1min) = 26.25 C
b. how much work is done?
to get .175 A in a 12 Ω resistor requires: ∆V = I•R = .175 • 12 = 2.1 V
W = q∆V = 26.25 C • 2.1 V = 55.1 J
c. how much power
P = W/t = 55.1 J ÷ 150 seconds = .368 Watts or P = I2R = >.175 •.175 •12 = .368 Watts
3. Recall the rules for adding resistance networks. We have three resistors: 3 Ω, 5 Ω and 8 Ω.
a. The resistors are placed in series with a 12 V battery
i. find the total resistance
3 + 5 + 8 = 16 Ω
ii. find the current
I = ∆V/R = 12 V ÷ 16 Ω = .75 A **This is the current in each and every one of the resistors as well as the current from and into the battery
iii. find the potential difference across each resistor
∆V = IR so for the 3Ω we have 3 • .75 = 2.25 V and for the 5Ω we have 5 • .75= 3.75 V and for the 8Ω we have 8 • .75 = 6 V. To check: 2.25 + 3.75 + 6 = 12 V -- the same as the battery.
iv. find the work done in 30 seconds
q = I•t = .75 • 30 = 22.5 C then W = q∆V = 22.5 C • 12 V = 270 J (50.625 J at the 3 Ω, 84.375 J at the 5 Ω and 135 J at the 8 Ω)
v. find the power to each resistor
P = W/t = 50.625/30 and 84.375/30 and 135/30 = 1.6875 W, 2.8125 W, 4.5 W total = 9 W or
P = I∆V = .75 • 2.25 = 1.6875 W and .75 • 3.75 = 2.8125 W and .75 • 6 = 4.5 W total = .75 • 12 = 9 W
b. The resistors are placed in parallel with a 12 V battery
i. find the total resistance
add as reciprocals: 1/R = 1/3 + 1/5 + 1/8 so R = 1.52 Ω
ii. find the current through each resister
I = ∆V/R = 12/3 = 4 A and 12/5 = 2.4 A and 12/8 = 1.5 A (1.5 + 2.4 + 4 = 7.9 A altogether or 12/1.52 = 7.9 A)
iii. find the potential difference across each resistor
In parallel, each resistor gets all 12 V
iv. find the work done in 30 seconds
q = I•t = 7.9 C/s • 30 s= 237 C
W = q∆V = 12 • 237 = 2844 J
v. find the power to each resistor
P = W/t = q∆V/t = It∆V/t = I∆V so we have 4 A • 12 V = 48 W, 2.4 A • 12 V = 28.8 W, 1.5 A • 12 V = 18 W
Resistance of a wire. We've toyed with the resistance equation, now we'll really work on it a bit. Resistivity is a property of conductors. Good conductors have low resistivity, poor conductors have high resistivity.
4. The resistivity of copper is 1.68 x 10-8 Ω-m. What is the resistance of a wire 12 meters long with a radius of .0005 m. Resistance is directly proportional to the length and inversely proportional to the cross-sectional area (think circle). The quotient is multiplied by the resistivity.
R = (rho)L/A = 1.68 x 10-8 Ω-m • 12 m ÷ (∏•.0005 •.0005) = .257 Ω
Coulomb's Law -- The force between two charged objects is directly proportional to the product of the magnitudes of the charges and inversely to the square of the distance between them.
5. The force between two charged objects is 2.0 N. If the magnitude of one of the charges is doubled, what will the new force be?
2.0 • 2 = 4 N
6. The force between two charged objects is 5.0 N. If the distance between the charges is halved, what will the new force be?
5.0 • 4 = 20 N
7. The force between two charged objects is .0048 N. If the magnitude of one charge is doubled, the magnitude of the other tripled and the distance between them is multiplied by 5, what will the new force be?
.0048 • 2 • 3 • (1/25) = .00115 N