What is the magnitude of charge on each sphere?

[AznBoi]

Homework Statement

Two small metallic spheres, each of mass 0.2g are suspendedas pednulums by light strings from a common point. The spheres are given the same eletric charge, and it is found that they come to equilibrim when each stringis at an angle of 5degrees with the vertical. If each string is 30 cm long, what is the magnitude of the charge on each sphere?

Homework Equations

F=kq1q2/r^2

The Attempt at a Solution

I figured I would need the distance between the spheres so I used x=.3m sin5 deg and multiplied x by 2 to get the distance (r) between them.

Do I add two force equations and put them equal to zero? What other equations do I need? If I try solving for q that way everything cancels out.

 

[wellcaffeinated]

Ok, so you have the distance, and you want the charge, so you need the force repelling these spheres.

 /\           y
O  O          |_x

Just consider one of these balls. You know that the electric force is keeping them apart, but there must be a force stopping them from flying apart in the x direction. Gravity points downwards (y direction), so it's not that.

The only other thing it could be is the string. The string's force (tension) can be broken up into components. One of these must be equal and opposite to gravity, and the other must be equal and opposite to the EM force, otherwise the spheres would not be stationary.

You know the gravitational force, so you know one of the components of the string. You can then find the other component of the string, which should give you what you need.

Hope this helps.

 

[AznBoi]

Hmm, I kind of get what you are saying. So I need to make the x component of the tension force equal to the outward eletric force?

I think I will be able to find tension because that's what the mass is for! I'll try it out, thanks!

 

[AznBoi]

I still didn't manage to get the correct answer.

Here's my work:

I found the (r) distance between the two metal spheres first.
sin5deg= \frac {5}{.03m}
2x=r=.0523m

I drew a FBD with <--- as F_e , ----> as T_x , Up vector as T_y and Down vector as the weight.

T_y=W=mg=(2*10^-4kg)(9.8)= 0.00196 N

Used sine/cosine to find T_x and got: T_x=1.7147*10^-4 N

Tx-F_e=0 so Tx=F_e

Here is where I think I've messed up but I'm not sure:

T_x=k_e\frac{2q}{r^2}

q=\frac{T_x*r^2}{2K_e}

and I solved for q and I got 2.6 x 10^-17

The correct answer is 7.2 nC. What am I doing wrong? Thanks!

 

[denverdoc]

It looks pretty at first glance, up until you get to 2Q, think that should be
q^2

 

[AznBoi]

It looks pretty at first glance, up until you get to 2Q, think that should be
q^2

ooh. but I thought 2q and q^2 are the same? I'll try it anyways, thanks!

 

[ranger]

ooh. but I thought 2q and q^2 are the same? I'll try it anyways, thanks!

Er, no.

q+q = 2q
q*q = q²

 

[AznBoi]

Woww! It works! Can you explain to me why I can't use 2q instead of q^2.

Edit: oh ok. xP I'll try to not make a mistake like that anymore.Thanks you guys! =D

 

[ranger]

Woww! It works! Can you explain to me why I can't use 2q instead of q^2.

Because coulomb's law needs the product of the two charges and not the sum.
k \frac{q1 \cdot q1}{r^2} \not= k \frac{q1+q1}{r^2}
q \cdot q \not = q+q<br />

When we have several of the same numbers, x, being multiplied together. It is more efficient to do xⁿ, where n is the number of x's we have.
If we have several of the same numbers, x, being added together. It is more efficient to do n*x, where n is the number of x's being added.

 

[AznBoi]

Because coulomb's law needs the product of the two charges and not the sum.
k \frac{q1 \cdot q1}{r^2} \not= k \frac{q1+q1}{r^2}
q \cdot q \not = q+q<br />

When we have several of the same numbers, x, being multiplied together. It is more efficient to do xⁿ, where n is the number of x's we have.
If we have several of the same numbers, x, being added together. It is more efficient to do n*x, where n is the number of x's being added.

Thanks for the awsome explanation! I completely get it now.