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Coulomb's Law-Electric Fields (help please!)

by moonlit
Tags: coulomb, fields, lawelectric
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moonlit
#1
Mar23-04, 01:15 PM
P: 58
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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Chen
#2
Mar23-04, 03:12 PM
Chen's Avatar
P: 1,004
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 seperated 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.
moonlit
#3
Mar23-04, 06:55 PM
P: 58
I understand problems one and three but the second problem I still don't understand. How do I know the radius of the sphere?

Chen
#4
Mar24-04, 01:03 AM
Chen's Avatar
P: 1,004
Coulomb's Law-Electric Fields (help please!)

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.


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