# Minimum work to transport electron?

1. Dec 6, 2015

### qlzlahs

1. The problem statement, all variables and given/known data
A charge Q = -820 nC is uniformly distributed on a ring of 2.4 m radius. A point charge q = +530 nC is fixed at the center of the ring. Points A and B are located on the axis of the ring, as shown in the figure. What is the minimum work that an external force must do to transport an electron from B to A?
(e = 1.60 × 10^-19 C, k = 1/4πε_0 = 8.99 × 10^9 N · m^2/C^2)

https://www.physicsforums.com/attachments/work-png.93021/?temp_hash=3fc763fb95a2d9ab71f3bf4a54a23c14

2. Relevant equations
V = (k*q)/(sqrt(R^2 + z^2))
work = (V_b - V_a)*q
work = (k*q_1*q_2)/r

3. The attempt at a solution
V_B = (9*10^9*530*10^(-9))/(3.2) = 1490.625 V
V_A = (9*10^9*530*10^(-9))/(1.8) = 2650 V
V_B - V_A = -1159.375 V

(V_B - V_A)*q, where q = 1.60*10^-19 C
(-1159.375)*(1.60*10^-19) = -1.855*10^-16 J

I'm not sure if I'm supposed to use -820 nC or 530 nC for the q value when calculating V_B or V_A.

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2. Dec 6, 2015

### TSny

Does the ring also contribute to the potential at point B?

3. Dec 6, 2015

### qlzlahs

I was assuming that if the ring contributes to the potential at point A, it would to point B as well.

4. Dec 6, 2015

### TSny

The same amount at both points?

5. Dec 6, 2015

### qlzlahs

So.. How do I know how much potential there is at point B?

6. Dec 6, 2015

### TSny

Look at your list of relevant equations.

7. Dec 6, 2015

### qlzlahs

Do I use the equation V = (k*q)/(sqrt(R^2 + z^2)) for both points A and B? With R = 2.4 and z = 1.8 for A, and z = 3.2 m for B?

8. Dec 7, 2015

### azizlwl

Coulomb's law, F=kQ1Q2/r^2
Work=Fr
Since F is not constant between A and B, we have to calculate based on small distances dr so that F is constant within it.
dW=F(r)dr

Edit
You have to apply Gauss law too for the ring.

Last edited: Dec 7, 2015
9. Dec 7, 2015

### Potatochip911

Why would he want to use Gauss's law for this question? Work is equal to change in potential energy since $\Delta K=0$, i.e. $W=\Delta U=q\Delta V$, in this case $q$ is the electron. Note that for continuous charge distributions(Like the ring of charge): $V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}$