Coulomb's Law-Electric Fields (help please)

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In summary, the three problems are:-The force of repulsion that two like charges exert on each other is 4.3 N-Two tiny conducting spheres are identical and carry charges of -20.1 C and +38.8 C. They are separated by a distance of 1.96 cm.-A charge of -3.06 C is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius = 0.134 m). The charges on the circle are -3.02 C at the position due north and +6.29 C at the position due east.-When the spheres touch each other the charges on each one equalizes, so
  • #1
moonlit
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I have 3 problems I'm stuck on for homework:

1) The force of repulsion that two like charges exert on each other is 4.3 N. What will the force be if the distance between the charges is increased to 9 times its original value?

Not sure what equation to use here to solve this one...

2) Two tiny conducting spheres are identical and carry charges of -20.1 C and +38.8 C. They are separated by a distance of 1.96 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 1.96 cm. Determine the magnitude of the force that each sphere now experiences.

What equation should I use here? The one I'm using below?

3) A charge of -3.06 C is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius = 0.134 m). The charges on the circle are -3.02 C at the position due north and +6.29 C at the position due east. What is (a) the magnitude and (b) direction of the net electrostatic force acting on the charge at the center? Specify the direction as an angle relative to due east.

For part A I used the equation Fe=Kq1q2/r^2
Fe=-3.06*10^9*(-3.02*10^-6)*(6.29*10^-6)=.058127148/(.134)^2=3.237199153
I know this is wrong, can anyone help me find my problem?
 
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  • #2
1) You know that the force between two charges is:
[tex]F_{ele} = K\frac{q_1q_2}{r^2}[/tex]
So to compare the force between and after the change, divide them:
[tex]\frac{F_{ele1}}{F_{ele2}} = \frac{K\frac{q_1q_2}{r_1^2}}{K\frac{q_1q_2}{r_2^2}} = \frac{r_2^2}{r_1^2}[/tex]
And if [itex]r_2 = 9r_1[/itex] then:
[tex]\frac{F_{ele1}}{F_{ele2}} = \frac{9^2r_1^2}{r_1^2} = 9^2 = 81[/tex]
So the answer is that the magnitude of the force will be 1/81 of its original value.

2) You can answer the first part of the question with the equation above for [itex]F_{ele}[/itex]. After they bring the spheres into contact, the charges on them move between the spheres so the after they are separated, the charge on each one is different than the charge specified in the beginning of the question. When the spheres touch each other the charges on each one equalizes, so if at first they had (-20.1c + 38.8c) = +18.7c, after they are separated they will each have half of that i.e +9.35c.

3) Calculate the force that the charges on the circle exert on the charge in the middle, one at a time. You will get the magnitude of the force that operates along the north-south line, and the magnitude of the force that operates along the east-west line. When you have those use pythagoras to find the net force. To find the angle, use any of the trigonometric functions on the right triangle that consists of the two individual forces and the net force.
 
  • #3
I understand problems one and three but the second problem I still don't understand. How do I know the radius of the sphere?
 
  • #4
You don't need to know that, the formula only requires the distance between the spheres which you have.

[tex]F_{ele} = K\frac{q_1q_2}{r^2}[/tex]
Just in case you are confused, r there is NOT the radius of the spheres. It is the distance between them.
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the electrostatic interaction between two 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.

2. How is Coulomb's Law related to electric fields?

Coulomb's Law is related to electric fields because it describes the force that charged particles exert on each other, and electric fields are the regions in space where this force is present. The direction of the electric field is in the direction of the force that would be experienced by a positive test charge placed in that field.

3. What is the equation for Coulomb's Law?

The equation for Coulomb's Law is F = k * (q1q2) / r2, where F is the force between two charged particles, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

4. What is the SI unit for electric field?

The SI unit for electric field is newtons per coulomb (N/C). This can also be written as volts per meter (V/m).

5. How does distance affect the strength of the electric field?

The strength of the electric field is inversely proportional to the square of the distance between the charged particles. This means that as the distance between the particles increases, the strength of the electric field decreases. In other words, the electric field becomes weaker as you move further away from the source of the field.

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