25: PSS 22.1 Electric Forces and Coulomb’s Law

INTRO:
Two charged particles, with charges q₁=q and q₂=4q, are located at a distance d=2.00cm apart on the x axis. A third charged particle, with charge q₃=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.
Find the position of charge 3 when q = 2.00nC .

The mathematical representation is based on Coulomb’s law:
F_1on2=F_2on1=K⋅|q₁|⋅|q₂| / r².

Part A: Which of the following sketches represents a possible configuration for this problem?

Enter the letter(s) indicating the correct graph(s) in alphabetical order. For example, if you think that A and B are correct, enter AB.

SOLUTION: 
begin by looking and analyzing the given problem
for the forces to be equivalent, q₃ must be significantly closer to q₁ than to q₂, because q₂ is much stronger
However, q₃ could technically be either between the charges or outside of them, as long as it obeys the rules we just stated. 
sketch A could work because it follows our definition
sketch B can't work because q₃ is much closer to q₂ than to q₁ so F_2on3 would be much greater than F_1on3
sketch C could work because it follows our definition
and sketch D couldn't work for the same reason sketch B can't. 

So, our final answer is AC

Part B:
Suppose charge 3 is placed in between the other two charges, as shown below. Use this diagram to show the force vectors representing all the forces acting on charge 3.

Draw each vector with its tail at charge 3. The orientation of your vectors will be graded. The length of your vectors will not be graded.

SOLUTION:
this one is relatively difficult to type out, but I will try to. 
both vectors point in toward charge 3
(q₁)⇒ (q₃) ⇐(q₂)

NOTES:
When charge 3 is placed in between the other two charges, the electric forces exerted on it have opposite directions. In this configuration, when charge 3 is at a point that satisfies the conditions given in the problem, the net force on it will be zero because F_2on3=F_1on3. If, instead, charge 3 is located anywhere to the left of charge 1, both the electric forces exerted on it point to the left. Thus, the net force on charge 3 will also be directed to the left. You may want to add this information to your pictorial representation.

You are trying to find the position on the x axis of charge 3. Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, label the position of charge 3 x₃. Because of the two possible configurations, your calculations should yield two values for x₃: a positive value x_3,r when charge 3 is to the right of charge 1 and between the two charges, and a negative value x_3,l when charge 3 is located to the left of charge 1 (that is, to the left of the origin).

Your final sketch should look like this:

Part C:
Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, find the two possible values x_3,r and x_3,ℓ for the position of charge 3.
Express your answers in centimeters to three significant figures, separated by a comma.

SOLUTION:
givens:
q₁ = q
q₂ = 4q
q₃ = q
d = 2.00 cm
x_q1 = 0 cm
x_q2 = 2.00 cm
q = 2.00 nC

next lets determine the forces
F_1on3= K⋅|q₁|⋅|q₃| / r² = K⋅|q|⋅|q| / (d_1-3)² = K⋅q² / (d_1-3)²
F_2on3= K⋅|q₂|⋅|q₃| / r² = K⋅|4q|⋅|q| / (d_3-2)² = K⋅4q² / (d_3-2)²

since the forces are supposed to equal each other in each case, we're able to solve for the variables in both cases

F_1on3= F_2on3
K⋅q² / x²=K⋅4q² / (2.00 cm - x)²
1/x² = 4/ [4.00 cm² - (4.00 cm⋅x) + x²]
[4.00 cm² - (4.00 cm⋅x) + x²]/x² = 4
4.00 cm²/x² - (4.00 cm⋅x)/x² +x²/x² - 4 = 0
4.00 cm²/x² - (4.00 cm)/x + 1 - 4 = 0
4.00 cm²/x² - (4.00 cm)/x -3 = 0
3=4.00 cm²/x² - (4.00 cm)/x
3x²=4.00 cm -4.00 cm⋅x
3x² + 4.00 cm⋅x - 4.00 cm = 0
plugging that into the quadratic equation we get 
x = -2 & x = 2/3 
x_3,r, x_3,ℓ = 0.667,-2.00 cm, cm

Part D:
The key equation for this problem is F_1on3=F_2on3. In the previous part, you solved it by applying Coulomb's law and obtaining a quadratic equation with two real-valued solutions for x₃. If, instead of writing an equation for x₃, you apply Coulomb's law and write a simpler equation in terms of the distances of charge 3 from charges 1 and 2, respectively, r₁₃ and r₂₃, which of the following relations would you get?

a) r₂₃=2r₁₃
b) r₂₃=½r₁₃
c) r₂₃=4r₁₃
d) r₂₃=¼r₁₃

SOLUTION:
If we just backtrack to some of the math we did in the last part, we find
K⋅q² / x²=K⋅4q² / (2.00 cm - x)²
x² is r₁₃ and (2.00 cm - x)² is r₂₃
so, 1 /r₁₃²=4/ r₂₃²
→r₂₃²=4r₁₃²
→r₂₃=2r₁₃
or option a)