Calculating Potential Difference in a Parallel-Plate Capacitor

[AznBoi]

Homework Statement

A small object with a mass of 350mg carries a charge of 30nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separacted by 4 cm. If the thread makes an angle of 15 degrees with the vertical, what is the potential difference between the plates?

The Attempt at a Solution

First, I drew a diagram. Since the object has a + charge, it would be more closer to the negative plate.

$$\Delta V=\frac{U_{E}}{q}$$

$$U_{E}=qEd$$ , $$E=\frac{kq}{r^2}$$

Now that I have all the required info to solve for the electrical potential energy, I solved it and put it into the equation: $$\Delta V=\frac{U_{E}}{q}$$

For the electrical potential energy I got: 2.023*10^-4J

For the answer (the potential difference) I got: 6742.5V

Does my method seem correct? This is an even problem so I have no way of checking my answer for accuracy. Thanks in adv.

 

[Hootenanny]

Do you use the mass of the object in your solution...

 

[AznBoi]

Do you use the mass of the object in your solution...

xD I knew that I left something out.. So does that mean my solution doesn't work? I don't know how to use mass in this equation. Do I have to use forces or something?? F=ma?

 

[Hootenanny]

xD I knew that I left something out.. So does that mean my solution doesn't work? I don't know how to use mass in this equation. Do I have to use forces or something?? F=ma?

Indeed, if the mass is suspended (i.e. stationary) what can you say about the weight of the mass in relation to the force exerted on it by the electric field?

 

[AznBoi]

Indeed, if the mass is suspended (i.e. stationary) what can you say about the weight of the mass in relation to the force exerted on it by the electric field?

Well you can use the weight to find the y component of tension and then solve for the x component of tension. Then you use F=ma with the electric force exerted opposite of the x component of tension?

So will it be:
$$\Sigma F_{x}=0$$

$$-T_{x}+F_{e}=0$$

$$F_{e}=T_{x}$$

Then you use this equation:
$$\Delta V=\frac{U_{E}}{q}$$

Instead, you solve the electrical potential energy with the electric force?

$$U_{E}=F_{e}*d$$

 

[Hootenanny]

Sounds spot on to me, but don't forget the electric force will also contribute towards the forces in the y direction

 

[AznBoi]

Sounds spot on to me, but don't forget the electric force will also contribute towards the forces in the y direction

Alrighty, thanks for your help! =] But I don't get how the electric force contributes to the forces in the y direction. Aren't the plates parallel and vertical? So wouldn't the electric force be horizontal, therefore only contribute to the x direction? I 'm probably wrong, but that's how I invision it to be.

 

[Hootenanny]

Alrighty, thanks for your help! =] But I don't get how the electric force contributes to the forces in the y direction. Aren't the plates parallel and vertical? So wouldn't the electric force be horizontal, therefore only contribute to the x direction? I 'm probably wrong, but that's how I invision it to be.

Oops, my bad, I misread the question, I though the plates were at an angle . You're right, and its time for me to take a break...