# Find the speed of the charge.

1. Sep 11, 2015

### Antonius

1. The problem statement, all variables and given/known data

Positive point charge (Q = XXX μC, mass m = XXX g) is fixed at point (Y2 cm,0). A second identical charge q is constrained to slide on a friction-less wire along the y-axis. Assume: the only force on q is the electrostatic force. If q starts at (0, YYY) and is released from rest, find its speed when it reaches (0,YYY cm), in m/s.

2. Relevant equations

W = ΔV * q

ΔV = -kQ (1/r2 - 1/r1) ----( this equation was already derived from integrating with limits A to B - ∫ E * dr )

W = ΔK

3. The attempt at a solution

I understand, sort of, that I need to find energy needed to move the charge for that distance (y2 - y1). In order to get that, I need to find ΔV so that I could multiply the answer to charge q and make it equal to [(1/2) (m*Vf^2)] to get the Vf.

I used ΔV = -kQ (1/r2 - 1/r1)
and I got ΔV = 104500. 5716 V. Used [W = ΔV * q] to get W. Then set W = 1/2 mvf^2. Found vf = 2.647 m/s

Is this an OK logic to solve this problem? :/
I need an advice regarding the formulas I am using... not the correct answer...

Thank you.

2. Sep 11, 2015

### TSny

You have the correct approach to the problem.

However, you might have a couple of sign errors. If W represents the work done on the charge q by the electric force, then check your notes to see if W = +ΔVq or W = -ΔVq.

Also, check if ΔV = +kQ(1/r2 - 1/r1) or ΔV = -kQ(1/r2 - 1/r1).

3. Sep 12, 2015

### Antonius

$$(\frac {qQ} {4 π ε_o}) \int_{a}^{b} \frac {1} {r^2} dx = (\frac {qQ} { 4πε_o}) ((\frac {-1} { b}) - (\frac {-1} {a})) = - (\frac {qQ} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})]$$

The work done is by the charge that has a fixed position on the x - axis, I am saying?

Also, the integration that I just did gives me $\ W$ ... but why do I have both $q$ and $Q$ involved in the equation?

My professor gave me an equation similar to the one that I have derived, but it only includes Q as a reference charge (Electric field created by point charge Q) and gives off almost exact result. To be precise it follows like this:

$$\Delta V = V_b - V_a = - \int_{a}^{b} E \cdot dr = - (\frac {Q} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})]$$

So I am predicting, to find the $W$ done by $Q$ on $q$ I have to use the one I derived (from Coulomb's Law) and include both $Q$ and $q$. Whereas to find the $\Delta V$ I need to use the second equation that my professor provided? Well, in fact, I can use the second equation and still find the $W$ by multiplying the result to the value of charge $q$ (the charge that Work is being done upon)

So, to conclude, one finds $W$ done by Q on q and another finds $\Delta V$ in the case of charge Q creating an electric field which is not constant.

Hoping that my thoughts do not have flaws...

P.S. Since $a$ is less than $b$ , the fraction will work out to be $negative$ there fore the $W$ done will be $positive$ if that's what you meant to emphasize on. Since the charges would naturally move away from each other, I believe $W$ must be positive in this case.

Last edited: Sep 12, 2015
4. Sep 12, 2015

### TSny

Yes, that all looks good! So, for $b>a$ the electric force does a positive amount of work on q as q moves from $r = a$ to $r = b$.

Because the electric force is proportional to $Qq$ according to Coulomb's law.

The integration of $\frac{1}{r^2}$ here will produce a minus sign that cancels the minus sign in front of the integral. So, the result should not have the minus sign out front. Thus $\Delta V = V_b - V_a = - \int_{a}^{b} E \cdot dr = +(\frac {Q} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})]$. If you think about it, this makes sense. Since $b > a$, this gives $\Delta V < 0$. But that's what you expect as you move farther away from Q.

Yes, but to get the signs to work out you need to use the correct relation $W = -q \Delta V$ rather than $W = +q \Delta V$. The general relation between the work done by a conservative force and the change in potential energy is $W = - \Delta U$, with a minus sign. For electrostatics, $U = qV$, so $W = - q\Delta V$

Yes, $W$ should be positive in this problem. This comes about because $W = -q \Delta V$ while $\Delta V < 0$.

5. Sep 12, 2015

### Antonius

Yep. It's all clear. Thank you :)