Coloumbs Law Find the Force

In summary, the problem involves finding the force on two point charges, one at the origin and one on the x-axis, using Coulomb's law and the Coulomb constant. The units of the charges need to be in coulombs, and the force can be calculated by plugging in the values into the equation (Kc * Q2 * Q1)/R^2. In part 2, one of the charges is removed, resulting in a negative answer or the same equation as the electric field of a point charge.
  • #1
sushi362
5
0

Homework Statement


Part 1:

A charge of 2 µC is at the origin, and a charge
of 6 µC is on the x-axis at x = 1 m.
Find the force on charge q2. The Colulomb
constant is 8.98755 × 10^9 N · m2/C2.

Answer in units of N

Part 2:

Find the force on q1.
Answer in units of N


Homework Equations



(Kc * Q2 * Q1)/R^2
Kc = 8.98x10^9

The Attempt at a Solution



So basically, for part one, i did..

(Kc * 2 * 6)/(1^2) = 1.07850E11
answer is positive so they should be repelling each other, but the program says its wrong, and I am not sure where to go for part 2.

Thanks.
 
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  • #2
When doing physics always, always mind your units, you should have put
-(Kc*2E-6*6E-6) / (1^2) = -0.108N

The second part should be the same answer, if not it will be the answer to Part 1 in its negated form (0.108N) or it will be basically the same equation with one of the point charges removed, i.e.

(Kc*q2) / (R^2)

But that is the expression for and electric field of a point charge, so Part 2 should just be the negative answer of part 1.
 
Last edited:
  • #3
JonDrew said:
When doing physics always, always mind your units, you should have put
(Kc*2E-6*6E-6) / (1^2) = 0.108N

The second part should be the same answer, if not it will be basically the same equation with one of the point charges removed, i.e. (Kc*q2) / (R^2) = F

What i don't understand is, my professor did a similar problem in class, except with 3 charges, and used the units in microcoloumbs and got the right answer? Or should i be doing all the problems in coloumbs instead?
 
  • #4
sushi362 said:
What i don't understand is, my professor did a similar problem in class, except with 3 charges, and used the units in microcoloumbs and got the right answer? Or should i be doing all the problems in coloumbs instead?
You should convert the charges to coulombs from micro coulombs. You can verify this by paying careful attention to the units, particularly the units of Coulomb's constant. If your teacher used micro coulombs, perhaps he had adjusted the constant beforehand; otherwise he couldn't have gotten the right answer.
 
  • #5


Part 1:

The force on q2 can be found using Coulomb's Law:

F = (Kc * q1 * q2)/r^2

Where:
Kc = Coulomb's constant = 8.98755 × 10^9 N · m2/C2
q1 = charge at the origin = 2 µC = 2 × 10^-6 C
q2 = charge on the x-axis = 6 µC = 6 × 10^-6 C
r = distance between the two charges = 1 m

Substituting the values in the equation, we get:

F = (8.98755 × 10^9 N · m2/C2 * 2 × 10^-6 C * 6 × 10^-6 C)/(1 m)^2
= 1.07850 × 10^11 N

Therefore, the force on q2 is 1.07850 × 10^11 N, in the direction away from q1, as expected.

Part 2:

Similarly, for the force on q1, we have:

F = (Kc * q1 * q2)/r^2

Where:
Kc = Coulomb's constant = 8.98755 × 10^9 N · m2/C2
q1 = charge at the origin = 2 µC = 2 × 10^-6 C
q2 = charge on the x-axis = 6 µC = 6 × 10^-6 C
r = distance between the two charges = 1 m

Substituting the values in the equation, we get:

F = (8.98755 × 10^9 N · m2/C2 * 2 × 10^-6 C * 6 × 10^-6 C)/(1 m)^2
= 1.07850 × 10^11 N

Therefore, the force on q1 is 1.07850 × 10^11 N, in the direction towards q2, as expected.
 

1. What is Coloumb's Law?

Coloumb's Law is a fundamental law of electromagnetism that describes the relationship between electric charges and the force between them. It states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

2. How do you calculate the force using Coloumb's Law?

The force between two charges can be calculated using the equation F = k(q1q2)/r2, where F is the force, k is the Coulomb's constant (9x109 Nm2/C2), q1 and q2 are the magnitudes of the two charges, and r is the distance between them.

3. What is the unit of measurement for the force calculated using Coloumb's Law?

The unit of measurement for the force calculated using Coloumb's Law is Newtons (N).

4. How does the distance between two charges affect the force between them?

The force between two charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases, and as the distance decreases, the force increases.

5. Can Coloumb's Law be applied to non-point charges?

Yes, Coloumb's Law can be applied to non-point charges by using the principle of superposition. This means that the force between two non-point charges can be calculated by considering each charge as a collection of point charges and summing up the individual forces between them.

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