How Far Must a Proton Be from a Charge to Balance Its Weight?

[map7s]

Homework Statement

A point charge q = -0.80 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on the proton to be exactly opposite to its weight? (Let the y-axis be vertical and the x-axis be horizontal.)

Homework Equations

F=(kq1 q2)/r^2

The Attempt at a Solution

I first tried drawing this out and it seems like it should be a pretty straightforward problem, but I seem to be having more trouble with it than I probably should. I know that F would have to equal zero, but if I do that then I can't solve for r.

 

[berkeman]

The other relevant equation you need to use it the force due to gravity. Assume that this experiment is being done in a vacuum chamber on the surface of the Earth. Draw a free body diagram assuming that the charge is physically held at the origin, and think about how you would balance the gravitational force and the Coloumb force on the proton.

 

[map7s]

So if the proton is going to be placed below the point charge, then the gravitational force and the Coulomb force would be going in opposite directions, so you would have F(coulomb)-g...correct?

 

[berkeman]

So if the proton is going to be placed below the point charge, then the gravitational force and the Coulomb force would be going in opposite directions, so you would have F(coulomb)-g...correct?

Correct concept, but a small oversight in your math. You are correct that you want to balance the forces, but "g" is the acceleration due to gravity. How do you write the force due to gravity on a mass m?

 

[map7s]

Would you use the equation for the gravitational force? F=(Gm1 m2)/r^2

 

[berkeman]

Would you use the equation for the gravitational force? F=(Gm1 m2)/r^2

You could, but that's more complicated than necessary. Just use the appropriate form of the general F=ma, where you substitute the value of "g", the acceleration of gravity at the surface of the Earth. Then balance your forces, and you're done!

 

[map7s]

So...would it work if I did F= coulomb - mg to take the weight into account?

 

[berkeman]

So...would it work if I did F= coulomb - mg to take the weight into account?

In equilibrium (nothing is moving -- it's balanced), the sum of the forces on the proton must add to zero.

I have to bail for a couple hours, but you are on the right track. Write the value of the two forces (careful about signs!), set the sum to zero, and solve.

 

[map7s]

Would the charge of the proton just be equal to one? mg=(kq1 q2)/r^2

 

[berkeman]

Would the charge of the proton just be equal to one? mg=(kq1 q2)/r^2

No. "Charge" is in units of Columbs in the mks system of units. Be sure to stay consistent in your use of units. What can you say about the charge of a proton compared to the charge of an electron? What is the charge of an electron in Columbs?

 

[map7s]

The absolute value of the charges of protons and electrons are the same...so would I just use the charge of the electron in the equation?

 

[map7s]

I just wanted to make sure that I was using the correct equation: mg=(kq1 q2)/r^2

 

[berkeman]

Okie dokie. That's the correct force equation. What did you get for the final answer?

 

[map7s]

My final answer was 8.31229692 x 10^24 m