Tough physics probLem??*10 points*?

Question

pLs someone help me solve these.i already answered #1 and 2...bUt i dont know how to soLve the rest.

http://s3.supload.com/free/Picture_251-20080728035628.jpg/view/

q1=4 X 10^-6 C

q2=-3 X 10^-6 C

q3=-2.12 X 10^-6 C

q4=-2.12 X 10^-6 C

*10 cm apart from each other

angle=135

1. what is the magnitude of force exerted on charge q1 by charge q2?

ans: 10.8 N

2. what is the magnitude of force exerted on charge q1 by charge q3?

ans: 7.63 N

3. refer to question 2. what is the horizontal component of force exerted on q1 by q3?

a. 0 N

b. 0.108 N

c. 3.77 N

d. 5.39 N

e. 7.63 N

f. 10.8 N

4. what is the resultant force exerted on charge q1 by the other three charges?

5. if q1 were replaced by a charge of -4 X 10^-6 C, what would be the resultant force exerted on it by the other three charges?

(choose your answers from the choices given in question 3)

thanks sOo much! 10 points fOr best answer!;)

Answer 1

OK, so you have the answers to the first two, although I think there's a rounding error for number 2, I get 7.62 N, but it may depend on the accuracy of the constants you have. (Exact value of Coulomb's constant is 8987551787.3681764).

For part three you have to find the horizontal componant of the force with a little geometry. Picture a right angled triangle, with one angle at 45 degrees, and the hypotenuse of length 7.62. You need to find the length of the side adjacent to the angle, and for that you need the cosine.

cos 45 = x / 7.62

x = 7.62 * cos 45

x = 5.39 N

For part four, you need to notice that the charges above and below q1 are equal and are at an equal angle and distance away, therefore there is no resultant vertical force acting on q1.

Horzontally though, you have just worked out the force on q1 by q3, and the value is exactly the same for the force acting on q1 by q4. Therefore the force pulling q1 to the left is:

5.39*2 = 10.78 N

Now you know the force pulling to the right because it was the very first thing you worked out, but what your answer is entirely depends upon the rounding you use. Since you were only given 1 significant figure for two of the values, you can only give your answer to 1 significant figure. Therefore the resultant force is 0 N.

However, treating the values of the charges as exact with no margin for error, the resultant force is 0.006713 N to the right.

For the last part of the question, if the value of q1 were flipped to be negative but of the same magnitude, all of the attractive forces would become repulsive forces, and therefore the resultant 0.006713 N force would instead act towards the left. Alternatively, since you can only give the resultant force to 1 significant figure (which is 0), flipping the charge makes no difference at all, since -0 is still 0.