Find the velocity of a point charge in a ring of electric field

[jisbon]

Homework Statement

Ring of radius 10cm and charge of 3$\mu C$ at the origin.
Ring axis aligned with z.
Electric potential along z is V=$\frac{kQ}{\sqrt{z^2+R^2}}$
Then another charge of -5$\mu C$ and mass of 0.1kg released from the origin at u = 7m/s in k direction.
Find the velocity of q at z=3m

Relevant Equations

$E = k\frac{\lambda a}{(x^2+a^2)^{3/2}}$

Okay, I am not even sure how to startr with this question. But here's my theory:

First I will need to the electric field produced by the ring using the formula:

$E = k\frac{\lambda a}{(x^2+a^2)^{3/2}}$

After finding out electric field produced by ring, am I supposed to find out the acceleration on the charge using $a=\frac{eq}{m}$ ? If so, the direction will be the k axis I assume?

If the above holds true, then I will proceed to use kinematics equation to solve for speed using $v^2=u^2+2as$ ? Since I have final displacement and acceleration (if I am correct)

 

[Delta2]

Your approach would be right if the electric field (and hence the acceleration) were constant along the z-axis, but they are not.

You don't need to calculate the electric field ,neither the acceleration. You are given the potential $V$ as function of $z$. Use that the work of electric field on a charge $q$ between two points A and B along the z-axis is $W_{AB}=(V(z_A)-V(z_B))q$ and also use the work-energy theorem.

 

[jisbon]

Your approach would be right if the electric field (and hence the acceleration) were constant along the z-axis, but they are not.

You don't need to calculate the electric field ,neither the acceleration. You are given the potential $V$ as function of $z$. Use that the work of electric field on a charge $q$ between two points A and B along the z-axis is $W_{AB}=(V(z_A)-V(z_B))q$ and also use the work-energy theorem.

I actually kind of understood what you meant, but my workings doesn't reflect it. Here's my workings:
$W_{AB}=(V(z_A)-V(z_B))q$
$W_{AB}=3*10^{-6}(\frac{kQ}{\sqrt{R^2+3^2}}-\frac{kQ}{\sqrt{R^2}}) = 3*10^{-6}(\frac{(9*10^9)(-5*10^{-6})}{\sqrt{0.1^2+3^2}}-\frac{(9*10^9)(-5*10^{-6})}{\sqrt{0.1^2}}) = 1.30502...$
Since change in kinetic energy = Work done
$1.30502... = \frac{1}{2}mv_{f}^2-\frac{1}{2}mv_{i}^2$
$1.30502... = \frac{1}{2}(0.1)v_{f}^2-\frac{1}{2}(0.1)7^2$
$v_{f} = 8.666...$
which is incorrect. Any ideas?

 

[Delta2]

There is a sign error in the work $W_{AB}$ and that's because you took as point A the point at z=3. Take as point A the point at z=0 and point B the point at z=3.

The formula for work is for movement from point A to point B.

 

[jisbon]

There is a sign error in the work $W_{AB}$ and that's because you took as point A the point at z=3. Take as point A the point at z=0 and point B the point at z=3.

The formula for work is for movement from point A to point B.

I got it :) 4.79m/s.
Thanks so much