1.  How far apart must two electrons be if the
force between them is 3 N?
F = k q1 q2 / d2
d = √ (9 x 109 N m2/C2)(-1.6 x 10-19 C)(-1.6 x 10-19 C)
3N
d = 8.76 x 10-15 m
Electric Force: What is the net Force on the middle charge?
+75 mC
1.5 m
+45 mC
1.5 m
-90 mC
Fon middle due to L = (9 * 109 N m2/C2) * (75 * 10-3 C * 45 * 10-3 C) / (1.5 m)2
Fon middle due to L = 1.35 * 107 N, positive, pushing the middle charge to
the R
Fon middle due to R = (9 * 109 N m2/C2) * (-90 * 10-3 C * 45 * 10-3 C) / (1.5 m)2
Fon middle due to R = -1.62 * 107 N, negative, pulling the middle charge to
the R
Since they’re both moving the middle charge to the R, you need to
combine them so that the resulting force is larger than either individual
force.
Net Force = 1.35 * 107 N - -1.62 * 107 N
Net Force = 2.97 * 107 N, moving the middle charge to the right
This is the method to solve any Force or E field problem with multiple charges!
2. What is the magnitude and direction of the
electric field, at a location shown in the image,
due to a -20 µC and a +60 µC charge?
-20
5 cm
X
+60
15 cm
You need to determine the E field at that test point from each of
the individual charges, and then determine how to combine them
based on the direction of the field lines at the test point, that
come from each charge
-20
5 cm
X
+60
15 cm
Edue to L = (9 * 109 N m2/C2) * (20 * 10-6 C)
(5 cm * 1 m / 100 cm)2
Edue to R = (9 * 109 N m2/C2) * (60 * 10-6 C)
(10 cm * 1 m / 100 cm)2
Notice we’ve ignored the signs on the charges when
calculating E
E due to L = 7.2 * 107 N/C, pointing to the L, in to the negative
charge
Edue to R = 5.4 * 107 N/C, pointing to the L, away from the
positive charge
-20
5 cm
X
+60
15 cm
E due to L = 7.2 * 107 N/C, pointing to the L, in to the negative
charge
Edue to R = 5.4 * 107 N/C, pointing to the L, away from the
positive charge
Net Field = Edue to L + Edue to R because they’re both pointing in
the same direction
Net Field = 12.6 * 107 N/C, pointing left
1. What is the current flowing through this circuit?
15 kΩ
+9 V
3.6 x
10-4
10 kΩ
V = I R, where R is the total R in series, or (15 + 10) kΩ
I = ( 9 V ) / (25 x 103 Ω)
or 3.6 x 10-4 A
2. What is the total resistance in this circuit?
49.47
50 Ω
+12 V
Total R is the R for resistors in parallel
Rtotal =
1
(1 / 50) + (1 / 4.7 x 103)
Or 49.47 Ω
4.7 kΩ
3. What is the current running through R2 in this
circuit?
50 Ω
0.0026 A
+12 V
4.7 kΩ
You can use V = IR, where you remember
V through each branch of a parallel circuit
is the same as the V of the power supply
I = 12 V / (4.7 x 103 Ω)
3. A current of 1 A flows in a wire. How many
electrons are flowing past any point in the wire
per second?
6.25 * 1018
I = 1 Amp = 1 C/sec = nq / t
1 C/sec * (1 ele / 1.6 x 10-19 C)
Or 6.25 * 1018 ele / sec