# Homework Help: Electric field strenght

1. Jun 15, 2010

### Zamze

1. The problem statement, all variables and given/known data
a positivly spheric charged ball is placed on an isolating tripod. its going to be used to measure the voltage between 2 paralell metaldiscs. trough a tiny hole in the lower metaldisc the ball reaches in to the homogen field. the ball and the isolated tripod are on top of a scale. the values on the scale are registered. after that the charge on the metal discs are turned off and the scale numbers are read again.

calculate the voltage between the 2 metaldiscs if the balls charge is 12nC (12*10^-9) and the difference in the scale measurements are 0.25g. the distance between the discs are 5cm.

2. Relevant equations

E(N/C)=F/Q
E=(k*Q)/r^2
E=F/Q

3. The attempt at a solution

i found out that E=(mg)/Q but this still didnt help me. (E here is for electric field strenghts)
so i tried k wich is 8.99*10^9.
(k*Q)/r^2 and got 43142 N/C for E.
then my E*d = 2157 volt,
so then i involved E=(mg)/Q
(E*Q)/9.82 and got m = 5.25*10^-5.
and since delta m is 0.25g i did my m + delta m and got 3.027*10^-4.

this i then used as U/d=F/Q with my new m value.
so (mg)/Q = 247709
this i multibplied with d, 0.05 and got 12385. so the closest thing i can come up to, to reach the answer 10kV is 12385-2157. but i wouldent exactly understand why :/.

Last edited: Jun 15, 2010
2. Jun 15, 2010

### Axiom17

Can you show your calculations? Then can see where your method is going wrong

3. Jun 15, 2010

### Zamze

i found out that E=(mg)/Q but this still didnt help me. (E here is for electric field strenghts)
so i tried k wich is 8.99*10^9.
(k*Q)/r^2 and got 43142 N/C for E.
then my E*d = 2157 volt,
so then i involved E=(mg)/Q
(E*Q)/9.82 and got m = 5.25*10^-5.
and since delta m is 0.25g i did my m + delta m and got 3.027*10^-4.

this i then used as U/d=F/Q with my new m value.
so (mg)/Q = 247709
this i multibplied with d, 0.05 and got 12385. so the closest thing i can come up to, to reach the answer 10kV is 12385-2157. but i wouldent exactly understand why :/.

4. Jun 15, 2010

### collinsmark

[Edit: previous comment here removed.]
Be careful with your equation for E. Does E = (k*Q)/r^2 apply to a system of parallel plates (or is that the equation for a point charge)? Perhaps you should check that. You can use Gauss' law to derive the formula if you don't have it written down somewhere.

(Hint: The electric field between the plates will be in terms of some other variables/constants such as Q, A (area), $\epsilon _0$, and/or σ. Leave your intermediate answer (for the electric field between the plates) in terms of these additional variables/constants for now. They may end up canceling out later before you reach your final answer. )

Last edited: Jun 15, 2010
5. Jun 16, 2010

### Zamze

Youre right that was for point charge and not for parallel plates. that makes it harder for me since i cant calculate out the electric field strenght now.

6. Jun 16, 2010

### collinsmark

Again, you can use Gauss' law to derive the electric field on your own. Show your work, if you're stuck. Alternately, you might be able to find the electric field formula for parallel plates in your text or notes (I recommend Guass' law though given the choice -- I hate remembering specific formulae, and Gauss' law is useful enough that one can derive many electric field equations on the fly, specific to the given configuration.) You can use Guass' law to get the electric field of a single plate. You can use superposition if you need to for parallel plates.

But like I said before, your electric field equation might end up a function of other unknown variables or constants, such as (a) area of the plates, (b) surface charge density of the plates, (c) total charge on the plates, and/or (d) electric constant $\epsilon_0$. But don't worry about that. Leave your electric field equation in terms of any of these variables/constants for now. They will go away later. You don't need to look for a specific 'number' yet.

Once you find an equation for the electric field, you can find an equation for the voltage between the plates, in terms of the same variables/constants.

Also, once you find the electric field equation, you can find an expression for the force on the ball, in terms of the same variables/constants.

Using the data from the scale, you'll have enough information such that with just a small bit of substitution, you can find the voltage between the plates as an actual number (any unknown variables/constants vanish in this step).

7. Jun 16, 2010

### Zamze

I havent learned Gauss law yet.
but what it comes down to i guess is E=F/Q and E=U/d
where E is the electric field strenght, F is the force and can be mg, Q is the charge of the ball. U is the voltage between the fields, and d is the distance between the fields.
ive tried with F/Q=u/d but no progress.

8. Jun 16, 2010

### collinsmark

Okay, I understand now. Those equations were given beforehand. It turns out that those equations will work too, since they are valid for this particular problem with parallel plates. And thus it turns out that this problem is much simpler than I originally suspected, since you don't need to derive those equations.

So you have F/Q = U/d. Solve for the voltage, U. Substitute in the other values given in the problem.

(Cautious hint: The scale's measurement difference was given to you already as a difference. So there is no need to calculate any absolutes [such as the mass of the ball or tripod]. It is this difference [as given] that is associated with the electric force, thus the voltage. But be careful, the scale's measurement was given in units of kilograms [as many scales do], which is a measusre of mass. You'll need to convert this measurement to units of Newtons, which is a measure of force.)

[Edit: By the way, I looked through original post and I think you made a miscalculation somewhere. Start over with your F/Q = U/d equation. All the other information you need is given in the problem statement -- just make sure you convert kg to N.]

Last edited: Jun 16, 2010
9. Jun 16, 2010

### Zamze

thx alot for the help, managed to solve it now =).
cheers!

10. Jun 16, 2010

Great!