Detector moving toward charged ring - rate of change of V

[Alan I]

Homework Statement

A thin ring (radius r = 1.41 cm) carries a charge Q = 8.57 pC distributed uniformly along its length. The ring lies in the y-z plane, so the axis through its center is the x-axis .

A small detector is moving along the positive x-axis toward the ring at velocity v = -0.543i mm/s. When the detector reaches the point (5.91 cm,0), at what rate does it measure the potential is changing, in V/m? The sign will indicate whether it is increasing or decreasing.

Homework Equations

V = kQ / (x²+R²)^1/2

The Attempt at a Solution

I honestly cannot even figure out now looking at my notes what I actually did as it was towards the morning when I did this but this is what I have:

V = (Q/2πε₀r²)*[(x²+r²)^1/2-x]

It seems I took dV/dt and got (Q/2πε₀r²)* x/(x²+r²)^1/2

and then plugging in values I got - 0.409

 

[haruspex]

I suggest you start again with a clear head. Your 'relevant equation' is right, but I cannot see how you got from there to your next equation. We cannot point out where you went wrong if you do not post your working.

 

[rude man]

Confusing question. The problem asks for rate of change of potential in V/m, suggesting what is sought is dV/dx evaluated at P. On the other hand, the velocity of P is also given, suggesting what is sought is dV/dt evaluated at P.
Take your pick I guess ...

 

[mfb]

They could also ask how the electric field (in V/m) is changing as function of time.
Pick one, or calculate more than one.

 

[mfb]

Yes, but then why did they give the speed? Something is odd, I just considered another option.

 

[rude man]

Yes, but then why did they give the speed? Something is odd, I just considered another option.

Sorry, I thought I had deleted my post. Yes, agreed, as I say, a confusing question!