Too many variables!

[DivideByZero]

This is a hw problem so please explain your process instead of telling me the whole answer, thanks!

 

 

There is a conductive ball hanging from a string in a electric plate.

 

As you can see its slightly tilted toward the negative end.

 

mass of ball: 0.04g

potential difference: 480V

separation of plates: 0.06m

angle of string hanging: 20 degrees

 

the question asks: What is the charge of the ball?

 

I drew a free body diagram of three forces.

one: force of tension

two: gravity/weight

three: force of charge pointing towards the right

 

I think I'm suppose to add up the three forces to equal zero.

F1 + F2 + F3 = 0.

 

I think F3 should use the formula ©(q1)(q2)/(r^2). Am I right?

If so then what is r???

 

I'm so lost in this problem please help guys!

[Klaynos]

If I were you I'd write down the force due to an electric field to start with...

 

You should probably be familiar with the lorentz force

 

http://en.wikipedia.org/wiki/Lorentz_force

 

Or at least a very simplified version of it for a charge in a uniform electric field...

[DivideByZero]

If I were you I'd write down the force due to an electric field to start with...

 

You should probably be familiar with the lorentz force

 

http://en.wikipedia.org/wiki/Lorentz_force

 

Or at least a very simplified version of it for a charge in a uniform electric field...

 

electric field:

 

E = F/q

 

E * q = F

 

(480)(q) = F

 

 

holy s*** thanks for that hint!

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Merged post follows:

Consecutive posts merged

OK so here is what I've done so far:

 

F_g = m*g = (0.00004kg)*(9.8) = 3.92e-4

 

F_e = E*q = (480)*(q)

 

F_t = m*g*cos(20) = 3.684e-4

 

F_g + F_e + F_t = 0

 

...

 

q = 1.58e-6

 

 

 

is this right?????

[NeonBlack]

electric field:

 

E = F/q

 

E * q = F

 

(480)(q) = F

 

No.

 

Potential difference (voltage) is not the same as electric field. You will need to use the potential difference and the geometry of the plates to calculate the electric field.

[DivideByZero]

No.

 

Potential difference (voltage) is not the same as electric field. You will need to use the potential difference and the geometry of the plates to calculate the electric field.

 

 

oh so i have to use E=V/d

and add the vectors...