How Does an Electron Move in the Field of a Uniformly Charged Ring?

[defineNormal]

ok here goes,

A uniformly charged ring has a radius equal to 10.0cm and a total charge of +12.0nC. An electron is placed on the ring's axis at a distance of 25.0cm from the center of the ring and is constrained to stay on the axis of the ring. The electron when it reaches the center of the ring?

ok so
Q=+12.0nC
r=0.10m
x=0.25m

here's what I got so far

dE=Ke*dq/r^2
r=(x^2+a^2)^(1/2)
<after integration>
=Ke^x/(x^2+a^2)^(3/2)
I can't figure out how to find Ke> Am I heading in the right direction?

 

[Andrew Mason]

I am not sure what the question is. Is it: "what is the speed of the electron when it passes through the centre of the ring"?

If so, try using electric potential. What is the potential energy at a distance d along the axis? What is the potential energy at the centre of the ring? What is the difference? That has to be the kinetic energy of the electron.

AM

 

[defineNormal]

woops, determine speed :)

 

[defineNormal]

Ok this is what I got Ke(Q/(x^2+a^2) <--- square root over the donominator.
Q= 12.0 nC
a= 25.0cm => 0.25m
r= 10.0cm => 0.10m

the answer given is 1.19x10^7, I keep getting 1.488x10^12, however I see where I wen't wrong because Ke=8.99x10^9 Nm^2/C^2, so this means I have to convert 12.0 nC to C, and I don't know how?

 

[OlderDan]

Ok this is what I got Ke(Q/(x^2+a^2) <--- square root over the donominator.
Q= 12.0 nC
a= 25.0cm => 0.25m
r= 10.0cm => 0.10m

the answer given is 1.19x10^7, I keep getting 1.488x10^12, however I see where I wen't wrong because Ke=8.99x10^9 Nm^2/C^2, so this means I have to convert 12.0 nC to C, and I don't know how?

I assume if it really is nC it means nanoCoulombs or 10^-9C

 

[defineNormal]

but then the exponants cancel each other out?

 

[OlderDan]

but then the exponants cancel each other out?

Did you use the charge of the electron? Your equation below is not correct. The RHS is just K, not Ke

Ke=8.99x10^9 Nm^2/C^2

Other than that, check your computations. You appear to have the correct distances and the correct PE.