Electric Field Question Solution: Simplified Approach for Calculating Force

[Icetray]

[SOLVED] Electric Field Question

Homework Statement

http://img293.imageshack.us/img293/1464/document001uf1.jpg

Homework Equations

$$\frac{Q}{4\eta\epsilon_{o}\\r^{2}}$$
Weight = (m)(g)

The Attempt at a Solution

I broke uo the forces into $$F_{r}$$ (the force of repulsion), the weight and normal contact force. I get stuck after that and would appreciate any helpful hints. (Ignore any pencil marks on the question)

Thanking you in advance.

 

[CompuChip]

In equilibrium, there is no net force on the charges.
Thus, the force in both the horizontal and the vertical directions should cancel.
Obviously, the force of repulsion works along the line connecting the forces, so it only has a horizontal component. Then you get two equations:
(sum of vertical forces) = 0
=> gravitational force + vertical component of normal force = 0

(sum of horizontal forces) = 0
=> force of repulsion + horizontal component of normal force = 0

You should check that I included all the relevant forces and find out what signs they all get (e.g. gravitational force has opposite sign to the vertical component of the normal force, such that you get $- m g + N_y = 0$).

 

[Icetray]

In equilibrium, there is no net force on the charges.
Thus, the force in both the horizontal and the vertical directions should cancel.
Obviously, the force of repulsion works along the line connecting the forces, so it only has a horizontal component. Then you get two equations:
(sum of vertical forces) = 0
=> gravitational force + vertical component of normal force = 0

(sum of horizontal forces) = 0
=> force of repulsion + horizontal component of normal force = 0

You should check that I included all the relevant forces and find out what signs they all get (e.g. gravitational force has opposite sign to the vertical component of the normal force, such that you get $- m g + N_y = 0$).

Thanks for your reply, but I'm not really getting what you mean. I was thinking that I'd have to do something like the below:

N sin θ = mg and Ncosθ = $$F_{r}/$$

I was thinking that I'd have to do is, find and equation for N and then sub that into he second equation to get a value for $$F_{r}/$$ and then from there find an equation for the value of q.

Or is this what you're trying to get at?

 

[CompuChip]

Yes, now you are working in the right direction: that is what I tried to explain.
First use
N sin θ = mg
to get the expression for N, then use
$$N \cos\theta = \frac{Q}{4\eta\epsilon_{o}\\r^{2}}$$
to get Q (or first get Fr and then get Q, if that makes you feel more comfortable)

 

[Icetray]

Yes, now you are working in the right direction: that is what I tried to explain.
First use
N sin θ = mg
to get the expression for N, then use
$$N \cos\theta = \frac{Q}{4\eta\epsilon_{o}\\r^{2}}$$
to get Q (or first get Fr and then get Q, if that makes you feel more comfortable)

Ah! Now I get it. Many many thanks CompuChip. I shall attempt the question again. (:

 

[Icetray]

Hi,

Can someone just check and confirm with me if what I have done is indeed correct?

Thanks.

http://img134.imageshack.us/img134/8331/document002aw0.jpg

 

[CompuChip]

Assuming you did the number crunching correct, the answer will be right. At least you have used the right way to solve it

 

[Icetray]

Awesome! (: Thanks!