From the handout used on Thursday:

1.
a) 8kQ^2/(5^3/2)a^2
b)E = kQ/4a^2 directed to the right
c) -9kQ/2a
d)very similar to the graph of y = x*e^(-x^2)
e)v = (9kQq/m)^(1/2)

2.
a) left (-), right (+) from E field lines
b) 100 V
c) 1.3 x 10^-10F
d) 8.0 x 10^-16N directed to the right
e) by conservation of energy, v = 4.2 x 10⁶m/s

3.
#1.
i. 480 ohms, .25 A
ii. 360 ohms, .33 A
#2.
The resistances are the same as above. Since they are in series, the current rhough both is 0.143 A.
#3. in order, moving down the list: 2, 1, 3, 4
d) parallel: 70 W, series, 17.2 W

4.
a) 20 V
b) Q = 3.0 x 10^-8C
c)
i. 30 V since current is zero!
ii. E = 0 inside any conductor.
iii. With 30 V over the two gaps, using V = Ed, E = 60,000 N/C

5.
a) t = L/v₀
b) a = Dv₀²/L²
c) E = mDv₀²/qL²
d) V = mD²v₀²/qL²

Electric Current exam tomorrow. You might also plan to meet with your partner to get an idea of what you will be trying during the break for the battery project.