- #1
benji
- 48
- 0
So I thought I understood this electricity stuff pretty well--I've read through the chapter once in full and skimmed over it a couple times. I've studied all the examples and still my problems aren't coming out correct. Here is one problem in particular that is giving me a lot of trouble:
What I did with this problem is draw a free-body diagram for the forces that act upon the sphere attatched to the string. I found the force of gravity by simply using F=mg, so I have the left side of the right triangle. Then I solved for the electrical force between the two spheres by using F=(k*q1*q2)/(r^2). That gives me the bottom of the right triangle. So shouldn't I just be able to use simple trig to find the hypotenuse of the triangle, thus finding tension on the string? I did this and my answer came out to be 0.615 N--the correct answer is 0.813 N. Where did I go wrong? I've checked and re-checked my math and I still can't figure out why. So that was part (b), for part (a) of the problem I just used trig--inverse tangent multiplied by the bottom side of the trianle over the tension. For this answer I'm WAY off--I get -2.3X10^-21 degrees when the correct answer is 15.4 degrees.
So I know I wrote a little novel here, but I could really use the help!
Thanks.
A small sperical insulator of mass 8.00X10^-2 kg and charge +0.600 mC is hung by a thin wire of negligible mass. A charge of -0.900 mC is held 0.150 m away from the spere and directly to the right of it, so the wire makes an angle [theta] with the vertical. Find (a) the angle [theta] and (b) the tension in the wire.
What I did with this problem is draw a free-body diagram for the forces that act upon the sphere attatched to the string. I found the force of gravity by simply using F=mg, so I have the left side of the right triangle. Then I solved for the electrical force between the two spheres by using F=(k*q1*q2)/(r^2). That gives me the bottom of the right triangle. So shouldn't I just be able to use simple trig to find the hypotenuse of the triangle, thus finding tension on the string? I did this and my answer came out to be 0.615 N--the correct answer is 0.813 N. Where did I go wrong? I've checked and re-checked my math and I still can't figure out why. So that was part (b), for part (a) of the problem I just used trig--inverse tangent multiplied by the bottom side of the trianle over the tension. For this answer I'm WAY off--I get -2.3X10^-21 degrees when the correct answer is 15.4 degrees.
So I know I wrote a little novel here, but I could really use the help!
Thanks.