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

by moonlit
Tags: coulomb, fields, lawelectric
 P: 1,004 1) You know that the force between two charges is: $$F_{ele} = K\frac{q_1q_2}{r^2}$$ So to compare the force between and after the change, divide them: $$\frac{F_{ele1}}{F_{ele2}} = \frac{K\frac{q_1q_2}{r_1^2}}{K\frac{q_1q_2}{r_2^2}} = \frac{r_2^2}{r_1^2}$$ And if $r_2 = 9r_1$ then: $$\frac{F_{ele1}}{F_{ele2}} = \frac{9^2r_1^2}{r_1^2} = 9^2 = 81$$ 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 $F_{ele}$. 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.
 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. $$F_{ele} = K\frac{q_1q_2}{r^2}$$ Just in case you are confused, r there is NOT the radius of the spheres. It is the distance between them.