Confusion about Coulomb's Law and 3 Point Charges

In summary, the conversation is about a problem involving three charges arranged in the xy-plane. The goal is to find the expression for the force on one of the charges, as well as the magnitude and direction of the force. The conversation includes references to Coulomb's law and the use of position vectors. The main point of confusion is regarding the algebraic manipulation of the equations, specifically in regards to the denominator and the use of position vectors.
  • #1
Roodles01
128
0

Homework Statement


2 questions regarding the answer I have been given for this problem. Attachments are the problem & relevant worked answer I disagree with.

Problem
Three charges are arranged in the xy-plane as shown in attachment. A charge Q is at the point A with (x, y) coordinates (–2a, 4a), a charge –Q is at the point B which is the origin (0, 0), and a charge 2Q is at point C with coordinates (3a, 4a).

Find an expression for the force FC on the charge at C due to the other two charges and hence determine the magnitude |FC| of the force and the unit vector that specifies the direction of FC


Homework Equations


Coulomb's law
F = q1 q2 / 4 ∏ ε0 r123 rhat12

F = ke q1 q2 / r123 * rhat

where ke = 1 / 4 ∏ ε0


The Attempt at a Solution


FCA = 1/4∏ε0 qC qA / rCA3 * rCA

FCA = ke 2Q Q / (5a)3 * (5a ex)
OK, but then goes to

FCA = ke Q2 / a2 * 2/53 * (5a ex)

probably simple algebra but can't see how (5a)3 goes to a2 &53

Next query is regarding
FCB = ke * 2Q * -Q / (5a)3 * (3aex + 4aey)

How can the denominator be (5a)3 too, surely this is √20
Please assure me (or otherwise) that it's not (5a)3

Thank you
 

Attachments

  • coulomb problem.JPG
    coulomb problem.JPG
    3.8 KB · Views: 506
  • coulomb problem 02.JPG
    coulomb problem 02.JPG
    15 KB · Views: 441
Last edited:
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  • #2
Roodles01 said:
probably simple algebra but can't see how (5a)3 goes to a2 &53

There's an "a" in the numerator which leaves a2 in the denominator.

Roodles01 said:
How can the denominator be (5a)3 too, surely this is √20

Remember that rCA is the vector from C to A and is not the position vector of A
 
  • #3
Thank you.
Yes, easy points to remedy.
 

Related to Confusion about Coulomb's Law and 3 Point Charges

What is Coulomb's law?

Coulomb's law is a fundamental law of physics that describes the electrostatic interaction between charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

What are point charges?

Point charges are idealized particles with a mathematical point location and no physical size. They are used in Coulomb's law to simplify calculations and represent charged particles that are far enough apart to be considered as single points.

How is the force between three point charges calculated?

The force between three point charges is calculated by using the principle of superposition, which states that the total force on a charged particle is the vector sum of the individual forces exerted by each of the other charged particles. This can be represented mathematically as F = k(q1q2/r12^2 + q1q3/r13^2 + q2q3/r23^2), where k is the Coulomb's constant, q1, q2, and q3 are the charges of the three particles, and r12, r13, and r23 are the distances between them.

What is the direction of the force between three point charges?

The direction of the force between three point charges is always along the line connecting the charges and is either attractive or repulsive, depending on the sign of the charges. Like charges (positive-positive or negative-negative) repel each other, while unlike charges (positive-negative) attract each other.

How does distance affect the force between three point charges?

According to Coulomb's law, the force between three point charges decreases as the distance between them increases. This is because the force is inversely proportional to the square of the distance between the charges. Therefore, as the distance increases, the force decreases exponentially.

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