1. Electric Force vs. Gravitational Force
K q1 q2
FE =
k = 9  109 N m2 / C2
d 2
G m1 m2
Fg =
G = 6.67  N m2 / kg2
d 2
Gravity is the dominant force when it comes to shaping galaxies and the like, but notice that k is about 20 orders of magnitude greater than G. Technically, they can’t be directly compared, since they have different units. The point is that a whole lot of mass is required to produce a significant force, but a relatively small amount of charge can overcome this, explaining how the electric force on a balloon can easily match the balloon’s weight. When dealing with high-charge, low-mass objects, such as protons & electrons, the force of gravity is negligible.

2. Electric Force Example
A proton and an electron are separated by 15 μm. They are released from rest. Our goal is to find the acceleration each undergoes at the instant of release.
Find the electric force on each particle.
Find the gravitational force on each particle. A proton’s mass is  kg, and an electron’s mass is 9.11  kg.
Find the net force on each and round appropriately. Note that the gravitational force is inconsequential here.
Find the acceleration on each particle.
Why couldn’t we use kinematics to find the time it would take the particles to collide?
1.024 × N
4.51 × N
1.024 × N
e-: × 1012 m/s2, p+: × 108 m/s2
d changes, so F changes, so a changes.
15 μm + -

3. System of Charges 767.2 N at 59.6 º N of W
Here four fixed charges are arranged in a rectangle.
Find Fnet on charge D.
Solution:
-16 µC
+25 µC A C
767.2 N at 59.6 º N of W
4 cm B D
+9 µC
3 cm
-7 µC
Link

4. Point of Equilibrium
Clearly, half way between two equal charges is a point of equilibrium, P, as shown on the left. (This means there is zero net force on any charge placed at P.) At no other point in space, even points equidistant between the two charges, will equilibrium occur.
Depicted on the right are two positive point charges, one with twice the charge of the other, separated by a distance d. In this case, P must be closer to q than 2 q since in order for their forces to be the same, we must be closer to the smaller charge. Since Coulomb’s law is nonlinear, we can’t assume that P is twice as close to the smaller charge. We’ll call this distance x and calculate it in terms of d.
x = ? +q P +q +q P
+2 q d

5. Point of Equilibrium k q q0 k (2 q) q0 = x 2 (d - x)2
Since P is the equilibrium point, no matter what charge is placed at P, there should be zero electric force on it. Thus an arbitrary “test charge” q0 (any size any sign) at P will feel a force due to q and an equal force due to 2 q. We compute each of these forces via Coulomb’s law:
+2 q d x +q P
The k’s, q’s, and q0’s cancel, the latter showing that the location of P is independent of the charge placed there. Cross multiplying we obtain:
k q q0
k (2 q) q0 =
x 2
(d - x)2
(d - x)2 = 2 x 2  d x d + x 2 = 2 x  x x d - d 2 = 0.

6. Point of Equilibrium -2 d  (2 d )2 - 4 (1) (-d 2 ) -2 d  8 d 2 x = =
From x x d - d 2 = 0, the quadratic formula yields:
-2 d  (2 d ) (1) (-d 2 )
-2 d  8 d 2
x = =
2 (1) 2
= -d  d Since x is a distance, we choose the positive root:
x = d ( )  0.41 d. Note that x < 0.5 d, as predicted.
Note that if the two charges had been the same, we would have started with (d - x)2 = x 2  d x d + x 2 = x 2  d x d = 0  d (d - 2 x ) = 0  x = d / 2, as predicted. This serves as a check on our reasoning.