What is the electric potential energy of a hydrogen atom in a circular orbit?

[t_n_p]

Homework Statement

In a model for the hydrogen atom, an electron is in circular orbit at mean distance Ro from the nucleus.

Sketch U as a function of separation distance r. Write an algebraic expression for the electric potential energy U for a model of this system, i.e. charges +e and -e separated by distance Ro. Be sure to state the reference you have used for zero potential energy.

The attempt at a solution

The only relevant (?) formula I could find is this one:
V=q/(4pi*εo*r)

How I can sketch the function or write the algebraic expression from that is beyond me though..

 

[rootX]

"charges +e and -e separated by distance Ro"
so, you know q, and you have the algebraic expresion.
"V=q/(4pi*εo*r)"
And the independent variable is R(or = x).

and put V on the y axis.

 

[t_n_p]

for q do I insert the charge on an electron, 1.60217646 × 10-19.

Edit: figured it out using this method...

http://img524.imageshack.us/img524/4633/potentialyp5.jpg

 

[t_n_p]

huh, so i forgot to square radius?

 

[learningphysics]

No no never mind... I totally messed. You did evernthing right tnp, I'm messing up here. sorry lol.

 

[t_n_p]

?
I did?

charges +e and -e, hence q1 = -1.6*10^-19 and q2 = 1.6*10^-19

 

[learningphysics]

?
I did?

charges +e and -e, hence q1 = -1.6*10^-19 and q2 = 1.6*10^-19

Yeah... that's what you did in the diagram... you used q, and Q... I didn't notice that...

 

[t_n_p]

Yeah... that's what you did in the diagram... you used q, and Q... I didn't notice that...

lol, had my confused there for a moment!

 

[learningphysics]

lol, had my confused there for a moment!

lol. sorry.

one thing though, I don't think a negative r makes sense here since r is the distance... the potential energy between a positive and negative particle is always negative...

 

[t_n_p]

So I wipe the negative x-axis. Thanks for pointing that out!

 

[t_n_p]

Got another Q,

Points A,B,C and D are at corners of a square of sides 2*2μm. Charges of +1nC and -1nC are placed at corners A and B respectively. What is the change in the electric potential from C to D.

A------B
| |
| |
C------D

 

[learningphysics]

Find the potential at D... find the potential at C... subtract the two.

 

[rootX]

I guess you don't need to do any calculations, just analyze the diagram carefully!

 

[t_n_p]

Find the potential at D... find the potential at C... subtract the two.

Using this formula?
http://img175.imageshack.us/img175/3082/formulall1.jpg

I guess you don't need to do any calculations, just analyze the diagram carefully!

not sure what you're getting at

 

[learningphysics]

Using this formula?
http://img175.imageshack.us/img175/3082/formulall1.jpg

If there were two different r's r1 and r2 for q1 and q2, then it would be right...

The potential due to a point charge at a point r away is kq/r or q/4(pi*e0*r)

What is the potential at C due to the charge at A?
What is the potential at C due to the charge at B?

 

[t_n_p]

Does the sign of the charge matter?

For potential at C due to charge at A, I get
(8.99*10^9)(1*10^-9)/(2*10^-6) = 4495000V
Quite a large figure!

For potential at C due to charge at B, I get
(8.99*10^9)(1*10^-9)/(2*root(2)*10^-6) = 3178445V

Total potential at c = 3178445V + 4495000V = 7673444V

 

[learningphysics]

Does the sign of the charge matter?

For potential at C due to charge at A, I get
(8.99*10^9)(1*10^-9)/(2*10^-6) = 4495000V
Quite a large figure!

For potential at C due to charge at B, I get
(8.99*10^9)(1*10^-9)/(2*root(2)*10^-6) = 3178445V

Total potential at c = 3178445V + 4495000V = 7673444V

Yes, sign matters... keep the sign of the charge.

 

[t_n_p]

Yes, sign matters... keep the sign of the charge.

In that case, total potential at C = 1316555V

and

total potential at D = -1316555V

I want the change in the electric potential from C to D, so I subtract C from D?

 

[learningphysics]

In that case, total potential at C = 1316555V

and

total potential at D = -1316555V

I want the change in the electric potential from C to D, so I subtract C from D?

Yup that's right.

 

[t_n_p]

cool,
change in electric potential from c to d = -1316555 -1316555 = -2633110V

Final part of the question,
A 3rd charge of +2nC is initially placed at C and then moved to D. What is the change in the potential energy of the system? Take care with sign

 

[learningphysics]

cool,
change in electric potential from c to d = -1316555 -1316555 = -2633110V

Final part of the question,
A 3rd charge of +2nC is initially placed at C and then moved to D. What is the change in the potential energy of the system? Take care with sign

Try to use your previous result to calculate this part.

You can also do it the more brute force way... find the potential energy of the system at the beginning... find the potential energy of the system at the end... and subtract.

 

[t_n_p]

So I should calculate the total potential at D from A, B and C
and then calculate the total potential at C from A, B and D and find the difference?

 

[learningphysics]

So I should calculate the total potential at D from A, B and C
and then calculate the total potential at C from A, B and D and find the difference?

No. You've calculated the potential difference between the points C and D due to the charges A and B... so when you take a charge and put it and C and then move it to D, you can calculate the change in energy of the system by q * potential difference (q is the charge that is moved)

charge*potential difference gives the change in energy. You don't have to worry about the potentials due to the new charge that is being moved...

Another way to solve the problem (you can do this to double check your answer) is to take calculate the potential energy (not potential) in the system before and after... So there are 3 charges... you have to add up the potential energies between Q1 and Q2... then Q2 and Q3... then Q1 and Q3... You have the formula for potential energy between two charges... $$\frac{kq_1q_2}{r}$$

Then do the same thing in the new configuration after the charge is moved to D.

Then subtract the two energies...

 

[t_n_p]

If I use energy = charge*potential difference
energy = (2*10^-9)*-2633110 ?

 

[learningphysics]

If I use energy = charge*potential difference
energy = (2*10^-9)*-2633110 ?

yup. that works here...

but it won't work in general... it works when you're just moving one charge... and if the charges that you used to calculate potentials all remain stationary.
If the charge at A moved to a different position, then the potentials get messed up...

 

[rootX]

Got another Q,

Points A,B,C and D are at corners of a square of sides 2*2μm. Charges of +1nC and -1nC are placed at corners A and B respectively. What is the change in the electric potential from C to D.

A------B
|...|
|...|
C------D

oops, I thought A and B have same charges.

You could have used this way, if they both had same charges:


A-----B
|...|
|...|
C------D

distance [AD] = distance [BC]

and distance [AC]=distance[BD]

hence, C and D have same potential, therefore potential difference between them is 0.

 

[rootX]

so, analyzing diagrams first can save a lot amount of time (and you don't even have to take care of signs..)

Like in this problem, you didn't have to calculate potential differences for both C and D
you could have calculated at C, and then multiply it be 2

 

[t_n_p]

yup. that works here...

but it won't work in general... it works when you're just moving one charge... and if the charges that you used to calculate potentials all remain stationary.
If the charge at A moved to a different position, then the potentials get messed up...

Is it really that easy!
Change in potential energy = (2*10^-9)*(-2633110) = -0.005J

Thanks to everyone who helped!