# Point Charge Check?

1. Feb 1, 2005

### Fused

Point Charge Check!?

Hi, I' did this problem but for some reason I'm not getting the right answer. If you could check to see where I made my mistake I'd really appreciate it. Thanks!

The problem:

A poitn charge (m=1 g) at the end of an insulating string of length 55 cm is observed to be in equilibrium in a uniform horizontal electric field of 12000 N/C where the pendulum's position is as shown in the figure, with the charge 12 cm above the lowest (vertical ) position. If the field points to the right, determine the magnitude and sign of the point charge.

I know the charge is postive because it is repelled by the electric force.

Because the system is in equilibirum, I know that
Ftx = Fe (force of the elctric field)
Fty = mg

Fty = mg
= (.001 kg) (9.8)
=.0098

To find Ftx we need the angle.

theta = invscos (43/53)
theta = 35.77
Fty = Ft cos (35.77)
.0098 = Ft cos (35.77)
Ft = 0.012078

Ftx = 0.012078 sin 35.77
Ftx = 0.00706
Fe = 0.00706

Fe = qE
0.00706 = q * 12000
q = 5.88 * 10 ^-7

Correct Answer is : 7.8 * 10 ^-7

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2. Feb 1, 2005

### Staff: Mentor

The only mistake I can see is that you got the angle slightly wrong. (I get 38.6 degrees.)

An easier way of solving this is to consider these equations:
$$T sin\theta = E q$$
$$T cos\theta = mg$$
You can then combine them to solve for q without needing to first find the tension (T):
$$q = mg tan\theta /E$$

My answer does not agree with the one given as correct.

3. Feb 1, 2005

### Fused

thanks docAl
it should be invcos(43/55) to get 38.57 degrees.. Hm. but still not the right answer.

4. Feb 1, 2005

### Staff: Mentor

I would say that the answer given is incorrect.

5. Feb 9, 2012

### SoundZombie

Re: Point Charge Check!?

I have the same problem right now, but where I havent had Physics or Math in a few years, I dont understand this part. I get how $$T sin\theta = E q$$, and I can get $$T cos\theta = mg$$, but how can these be combined to solve for q? I feel like Im missing something between the two equations and the final equation solving for q.

The best I can figure out is because $\frac{T sin\theta}{Eq}$=1, and $\frac{T cos\theta}{mg}$=1, we can do ($\frac{T sin\theta}{Eq}$)*($\frac{T cos\theta}{mg}$)=1, then solve for q?

Last edited: Feb 9, 2012
6. Feb 9, 2012

### Staff: Mentor

Re: Point Charge Check!?

Divide one equation by the other:
$$\frac{Tsin(\theta)}{Tcos(\theta)} = \frac{Eq}{mg}$$
then simplify by the obvious steps

7. Feb 13, 2012

### SoundZombie

Re: Point Charge Check!?

I just dont follow the reasoning as to why this is possible. I realize its the easier way to solve for q, but why is this possible?

8. Feb 13, 2012

### Staff: Mentor

Re: Point Charge Check!?

You can always divide both sides of a given equation by equal values (provided you don't divide by zero!). And the thing about equations is, one side is always equal to the other. So dividing one equation by another in the way shown above is both correct and expedient

9. Feb 13, 2012

### SoundZombie

Re: Point Charge Check!?

Thanks for the help and explanation!