Finding Electric Field and Work at a Given Distance? Need help with Problem 1

In summary: Q=Ze?No, because the potential at each radius is the same and Ze is just the charge of a single electron.No, because the potential at each radius is the same and Ze is just the charge of a single electron.
  • #1
Adeel Ahmad
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Need help with Problem 1

Homework Equations


E = kq/r2
V = -W/q
V = kq/r

The Attempt at a Solution


1a. dE = kdq/rRSUP]2[/SUP]
dq = p ds
dE = kpds/R2
Not sure if I am right up to this point and don't know how to proceed.

1b. I think the electric field would be zero since the enclosed charge cancels out. Not sure on this either.

1c. V = kq/r so V = kZe/a
W = -kZ2e2/a

1d. Not sure how to approach

Any help would be appreciated!
upload_2016-5-6_21-57-50.png
 
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  • #2
Assume the E field is radial and constant in magnitude at a given radius, ##r##, from the center. This really follows from the symmetry of the charge distribution. What is the total charge enclosed inside our sphere of radius ##r##? Try looking at Gauss's law.
 
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  • #3
Paul Colby said:
Assume the E field is radial and constant in magnitude at a given radius, rrr, from the center. This really follows from the symmetry of the charge distribution. What is the total charge enclosed inside our sphere of radius rrr? Try looking at Gauss's law.
So then from Gauss's law, E = q/εA so E=Ze/επR2
Would that be correct for R>r.
And in the case of r>R would it be zero, since for q it would cancel out to zero?
 
  • #4
Yes, so ##q(r)## is a function of ##r## which gives ##E(r)##. Just write out ##q(r)## and the rest follows.
 
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  • #5
Paul Colby said:
Yes, so q(r)q(r)q(r) is a function of rrr which gives E(r)E(r)E(r). Just write out q(r)q(r)q(r) and the rest follows.
Awesome, so when r>R, its zero. Thanks! Can you assist me in telling me if I am right for part c please? I got V = kq/r so V = kZe/a and W = -kZ2e2/a and can you guide me through part d please?
 
  • #6
Okay, the expression for the potential energy looks in order. Since the work needed has to equal the change in potential energy why does your expression for work have ##Z^2e^2## in it while the potential has just ##Ze##?

On d I'd point out that the potential energy of the electron is the sum of two parts. One due to the nucleus and one due to the cloud. Look at the signs of these.
 
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  • #7
Paul Colby said:
Okay, the expression for the potential energy looks in order. Since the work needed has to equal the change in potential energy why does your expression for work have ##Z^2e^2## in it while the potential has just ##Ze##?

On d I'd point out that the potential energy of the electron is the sum of two parts. One due to the nucleus and one due to the cloud. Look at the signs of these.
So I got the electric potential, V, to be kZe/a and plugged that into the equation for work which is W = -Vq which is where I got the -kZ2e2/a. So you're saying this is not correct?
 
  • #8
It boils down to what the symbols mean and what conventions one are using. Typically ##Z## is the integer number of charges in the nucleus and is the "Atomic Number". ##e## is the unit of charge or some ##1.6\times 10^{-19}## coulombs. ##k## is coulomb's constant. Okay, the electrostatic potential is ##V = kZe/a## and is in Joules/Coulomb (using MKS units). The potential energy is the ## V *(-e) ## so the answer should be ##-kZe^2/a## because the charge of the electron is ##-e##.
 
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  • #9
Paul Colby said:
It boils down to what the symbols mean and what conventions one are using. Typically ##Z## is the integer number of charges in the nucleus and is the "Atomic Number". ##e## is the unit of charge or some ##1.6\times 10^{-19}## coulombs. ##k## is coulomb's constant. Okay, the electrostatic potential is ##V = kZe/a## and is in Joules/Coulomb (using MKS units). The potential energy is the ## V *(-e) ## so the answer should be ##-kZe^2/a## because the charge of the electron is ##-e##.
But the q that is given is Q = Ze, so wouldn't you plug that in for W = -Vq which is where you would get -kZ2e2/a
 
  • #10
I may have lost the bubble on which question is being discussed, if so sorry. Problem c is what being asked about, correct? If so q is the charge of a single electron and not the cloud. ##-Ze## is the total charge of the electron cloud which has ##Z## electrons. Problem c doesn't have an electron cloud by my reading.
 
  • #11
Nope, the word "the" nucleus likely implies the cloud is there, my bad. There is a potential at each ##r## so you would need ##V(r)##. This would be multiplied by the electron charge which is ##-e## not ##-Ze##.
 
  • #12
Wouldn't it be multiplied by Ze since the electron cloud is there since Q is given to be +Ze?
 
  • #13
Additionally, the uniform charge density is also given and I haven't used it in any of the solutions thus far. Isn't the uniform charge density to be used in order to calculate the electric field?
 
  • #14
It's really important to ask oneself what the symbols mean. ##V## in the problem is always the electrostatic potential. This is the amount of work that needs to be done to move a charge, ##q##, to infinity (which is V=0 by the usual convention). If one doubles the charge it takes twice the work. Your answer would require 4 times the work because ##(2Z)^2=4Z^2## so something has to be wrong with your answer.
 
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  • #15
Ohhhh, ok I understand what you mean now. Thank you! But a last concern is that the uniform charge density is given in the problem but it's not used in any of the solutions. Is that fine?
 
  • #16
On problem c if we assume the electron cloud then at ##r=a## where the electron starts, the electron is starting at a lower potential because the cloud charge between the nucleus and the electron orbit "shields" the nuclear charge making it look smaller by the amount of charge in between. An extreme case if ##a > R##, ##R## being the atomic radius, it would take no energy to release the electron. My guess is that the electron cloud should be included and that one should calculate ##V(r)## and set ##r=a##. Your answer (less the additional factor of ##Z##) is correct assuming no electron cloud.
 
  • #17
So what would you get for the work as a final answer in part c?
 
  • #18
Well, The potential of the bare nucleus is,

##V_N(r) = k\frac{Ze}{r}##

The potential of the cloud is harder. The ##V## field due to the uniform charge is,

##V_C(r) = k\frac{Q(r)}{r}##

where ##Q(r)## is the total electron charge enclosed in the sphere of radius ##r##. That would be

##Q(r) = -Ze\frac{r^3}{R^3}## for ##r < R##

and

##Q(r) = -Ze## for ## r \ge R##

So adding these gives

##V(r) = V_N(r) + V_C(r) = kZe(1 - \frac{r^3}{R^3})\frac{1}{r}##

for ##r < R## and 0 outside ##R##. This is if I haven't dropped a factor so I'd check my answer carefully. Hope this helps
 
  • #19
Oh, and you'd have to evaluate at ##r=a##
 

1. What is an electric field?

An electric field is a region in which an electric charge experiences a force. It is created by other charges and can either attract or repel charged particles.

2. How is electric field strength calculated?

Electric field strength is calculated by dividing the force experienced by a charge by the magnitude of the charge itself. It is measured in newtons per coulomb (N/C).

3. What is the relationship between electric field and potential energy?

Electric field and potential energy are related by the equation U = qV, where U is potential energy, q is the charge, and V is the potential difference. This equation shows that electric field and potential energy are directly proportional.

4. How is work related to electric field?

Work is related to electric field through the equation W = -ΔU, where W is work, U is potential energy, and ΔU is the change in potential energy. This equation shows that work is done when there is a change in potential energy, which is caused by an electric field.

5. What is the difference between electric field and electric potential?

Electric field and electric potential are related but not the same. Electric field is a vector quantity that represents the force per unit charge, while electric potential is a scalar quantity that represents the potential energy per unit charge. In other words, electric field tells us the direction and strength of the force, while electric potential tells us the amount of potential energy a charge has at a certain point in the field.

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