How Fast Does the -8.051 μC Charge Travel to Position A?

[loganhey]

Homework Statement

A charge 36.41 mC is placed to the left of another charge 79.73 mC on a plane, as shown in the figure. Another charge of -8.051 μC and mass 14.31 g (depicted as a blue sphere) is placed at rest at a distance 16.34 cm above the right-most charge and released. What is the speed of the -8.051 μC charge if it reaches the gray position labeled A?

Homework Equations

E=F/q

The Attempt at a Solution

I actually have no idea

 

[SammyS]

Homework Statement

A charge 36.41 mC is placed to the left of another charge 79.73 mC on a plane, as shown in the figure. Another charge of -8.051 μC and mass 14.31 g (depicted as a blue sphere) is placed at rest at a distance 16.34 cm above the right-most charge and released. What is the speed of the -8.051 μC charge if it reaches the gray position labeled A?

Homework Equations

E=F/q

The Attempt at a Solution

I actually have no idea

Hello loganhey. Welcome to PF !

You will have to include a figure, or else do a much better job of explaining the given configuration, or both.

Also, you need to show an attempt at solving the problem, or at least show a reasonable attempt at understanding what you need to do to solve it.

 

[loganhey]

This is a picture of the question. It's an online assignment.

http://i298.photobucket.com/albums/mm258/wolfboy2712/Capture-1_zps7d609984.png


And THIS is a scan of my work on the question

http://i298.photobucket.com/albums/mm258/wolfboy2712/e058dbf8-94c4-4a5b-940e-b64df37fe12c_zpsf6ce9faa.jpg

I've gotten answers of 3004, 1611, 1323, and 94.2, all in m/s, none of them were correct. After the first wrong answer, I was given a formula equation the initial potential energy, the total energy at that point, (easy enough to derive, just calculating it for the point at which it rests initially with reference to the charges, and then equating that with the sum of all energy at point A). And it gave me the answer 94.2. I am dead certain of my calculations, so what is the answer that someone on here gets and how did you arrive at it if you could be so kind as to explain?

 

[SammyS]

This is a picture of the question. It's an online assignment.

http://i298.photobucket.com/albums/mm258/wolfboy2712/Capture-1_zps7d609984.png
http://i298.photobucket.com/albums/mm258/wolfboy2712/e058dbf8-94c4-4a5b-940e-b64df37fe12c_zpsf6ce9faa.jpg

I've gotten answers of 3004, 1611, 1323, and 94.2, all in m/s, none of them were correct. After the first wrong answer, I was given a formula equation the initial potential energy, the total energy at that point, (easy enough to derive, just calculating it for the point at which it rests initially with reference to the charges, and then equating that with the sum of all energy at point A). And it gave me the answer 94.2. I am dead certain of my calculations, so what is the answer that someone on here gets and how did you arrive at it if you could be so kind as to explain?

Here at PF, it's our philosophy to help you solve the problem.

What is the potential energy of the "Blue" charge in its initial position?

What is the kinetic energy of the "Blue" charge in its initial position?

What is the potential energy of the "Blue" charge when it gets to point A ?

 

[loganhey]

What is the potential energy of the "Blue" charge in its initial position?

Using the formula (Q₁q)/(4∏ε₀r₁) + (Q₂q)/(4∏ε₀r₂) + (1/2)mv²
(since there is no kinetic energy in the initial position) I can ignore the "(1/2)mv²"

Using -8.051*10^-6C as "q"; 0.03641C as Q₁; 0.07973C as Q₂; 0.3922m as r₂; 0.1634m as r₂

I get the answer -42044 J as my initial potential energy, and since there is no kinetic initially, -42044 J is my total energy.

What is the kinetic energy of the "Blue" charge in its initial position?
There is none.


What is the potential energy of the "Blue" charge when it gets to point A ?
Using the same formula as before, but this time with r₁ = 0.1732; r₂ = 0.1846
I get -46485 J as my potential energy at point A. Setting my initial total energy (-42044) equal to -46485 + (1/2)mv²: the kinetic energy I get is 4441 J. I multiply by 2, divide by the mass in kg (0.01431) and then square root that number (620684) to get my answer in m/s, 787.83. That is wrong according to the website.

 

[SammyS]

What is the potential energy of the "Blue" charge in its initial position?

Using the formula (Q₁q)/(4∏ε₀r₁) + (Q₂q)/(4∏ε₀r₂) + (1/2)mv²
(since there is no kinetic energy in the initial position) I can ignore the "(1/2)mv²"

Using -8.051*10^-6C as "q"; 0.03641C as Q₁; 0.07973C as Q₂; 0.3922m as r₂; 0.1634m as r₂

I get the answer -42044 J as my initial potential energy, and since there is no kinetic initially, -42044 J is my total energy.

What is the kinetic energy of the "Blue" charge in its initial position?
There is none.


What is the potential energy of the "Blue" charge when it gets to point A ?
Using the same formula as before, but this time with r₁ = 0.1732; r₂ = 0.1846
I get -46485 J as my potential energy at point A. Setting my initial total energy (-42044) equal to -46485 + (1/2)mv²: the kinetic energy I get is 4441 J. I multiply by 2, divide by the mass in kg (0.01431) and then square root that number (620684) to get my answer in m/s, 787.83. That is wrong according to the website.

That looks like the right procedure.

Does the website care about significant digits?

 

[loganhey]

No, it goes by a percentage error, as long as your answer is within 2% of the answer given (and they go by 3 SF standard, regardless of what the true SF is) you'll get it right.

 

[haruspex]

Maybe not quite accurate enough. I get some values up to 0.5% different for individual the PEs, but when you take the difference that same absolute error suddenly becomes about 5%. My final speed is only 2% different though, so I'm surprised your answer is not accepted.
Anyway, it is a good idea to track how errors propagate. If you take the difference of two numbers that only differ by 10% then your error %age jumps up by a factor of 10.