Question

"A wall has a negative charge distribution producing a uniform horizontal electric field. A ball of mass .01kg, carrying charge -80μC, is suspended by an uncharged, nonconducting thread, which is .3m long. The thread is attached to the wall and the ball hangs in equilibrium, as shown above (see the Twiddla.com link below), in the electric and gravitational fields. The electric force on the ball has a magnitude of .032N.
c) Determine the perpendicular distance from the wall to the center of the ball."

I am stuck on how to calculate this distance, with the given information!

Answer 1

The diagram: http://www.twiddla.com/1504676

Any and all help is greatly appreciated! :)

Answer 2

does the ball have a charge?

Answer 3

wait, it does lol

Answer 4

There is a formula that you can use to solve this. It's coulomb's law
F = kQ_1Q_2/r^2

Answer 5

k is a constant and it equals 8.987 x 10^9 Nm^2C^-2

Answer 6

I thought about that one, but what do I use as charge of the wall? The second q? :/

Answer 7

ok... then there might be another way. How is the ball hanged? Is there some angle?

Answer 8

Yes, there is an angle, because the ball, attached to the string, is pushed away from the wall, but it doesn't give the angle. It just says is pushed away.

Answer 9

I'm guessing that this is the diagram

Answer 10

Correct, I also have a link to a free body diagram above:)

Answer 11

ok, then you can find the angle. You can find the weight of the ball, and from there, you can get the length of hypotenuse

Answer 12

use sin(theta) = 0.032(force on ball)/hypotenuse. Then find the angle

Answer 13

after you get the angle, finding the distance is easy-peasy, because you can use trig to get the distance

sin(theta) = distance/length of rope, rearrange it to get you answer :)

Answer 14

theta = sin_(.032/.3)
= 6.123198866
= 6.12
This is what you meant, right? Use the string length as the "hypotenuse"?

Answer 15

Does that work, to mix forces and distances... ?

Answer 16

no no.... it does not work that way

Answer 17

you don't know the hypotenuse yet but you can solve for it using the both the weight and the force from the wall. so t^2 = weight^2 + force^2

Answer 18

use that t, plug it in the equation, and you should get 18.1 degrees

Answer 19

alternatively, I guess you can use tan or cos, depending on your taste :P

Answer 20

Oh, gotta, so the "y" is Fg on the ball... *doing math...

Answer 21

i got 18.1 degree for the angle...

Answer 22

Whoops, it's because I didn't orient theta correctly, to the placement of x and y.

Answer 23

sin (theta) = 0.032N/0.103N = 18.1

Answer 24

I got 17.7, but that could be rounding on one of our ends.

Answer 25

d = .0914327181
= .091m ?

Answer 26

You're satisfied with this problem?

Answer 27

I believe I know what to do... I did what alias said, and got reasonable numbers (posted above). It makes sense, and I see how to use forces to find the angle, which can be the applied to a known Langton using trig functions to find the unknown distance. :) if you're asking do I need more help, I don't think so:)

Answer 28

*a known length... Sorry!

Answer 29

0.093 is what i have, so you're close!!!

Answer 30

That could very well be rounding, or calculator differences, so yes. I'd say I understand the concept... It's not a conceptual error causing that inequality in our answers. Sweet! Thank you so very much for your help!

Answer 31

you're welcome :)