Distance where charged particle reaches half its final speed

[cajunchrisbu]

hey here is one i am kinda stuck on, the homework is due in a few hours and its my last problem. (any help would be appreciated!)
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A charge of 3.93 μC is held fixed at the origin. A second charge of 3.93 μC is released from rest at the position (1.15 m, 0.550 m).

Enter scientific notation as 1.23E4.

(a)
If the mass of the second charge is 3.33 g, what is its speed when it moves infinitely far from the origin?
----i found this number to be 8.10 m/s which is correct



(b) (so this is the part i need help with...)
At what distance from the origin does the 3.93-μC charge attain half the speed it will have at infinity?

im probably just having a brain fart but i can't seem to put it together...

 

[Dick]

If you put the first problem together ok you set the potential energy difference between r=1.15m and r=infinity equal to (1/2)mv^2. Right? Now set the difference between 1.15m and an unknown r equal to (1/2)m(v/2)^2 and solve for r. Right?

 

[cajunchrisbu]

for the first part i found the radius to the second charge from Pythagorean therm and used

k(qq)/r = (1/2)mv^2 and solved for v

so I am not understanding what youre saying to do because that is not how i did the first part...

if you could please type out the equations you are talking about so i can see them...

thanks

 

[Dick]

Ok. Yes, the radius for the first part is not 1.15m. Got it. Ok, so let r0 be the radius from the first part and v the velocity from the first part. Then solve k(qq)*(1/r0-1/r)=(1/2)m(v/2)^2. It's exactly the same idea.

 

[cajunchrisbu]

nice thanks that was exactly it i just wasnt seeing the front part of that equation...

i submitted the correct answer with 30 seconds to spare so i have a 100% in that class now...woot

thanks

 

[Dick]

nice thanks that was exactly it i just wasnt seeing the front part of that equation...

i submitted the correct answer with 30 seconds to spare so i have a 100% in that class now...woot

thanks

Congratulations! But it's the same thing you already did right? Just a potential difference.