KE and speed of a sphere near a fixed charge

  • Thread starter jbarnhart5
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  • #1
jbarnhart5
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Homework Statement



A tiny sphere of mass 8.60 µg and charge −2.80 nC is initially at a distance of 1.44 µm from a fixed charge of +8.53 nC.

(a) If the 8.60-µg sphere is released from rest, find its kinetic energy when it is 0.500 µm from the fixed charge.

(b) If the 8.60-µg sphere is released from rest, find its speed when it is 0.500 µm from the fixed charge.

Homework Equations



At least what I believe are relevant equations: PE1 + KE1 = PE2 + KE2
PE = k(q1q2/r)
KE = mv2/2

The Attempt at a Solution



The sphere is released at rest, so KE1 = 0 --> KE2 = PE2 - PE1 = ΔPE

ΔPE = ke(q1q2/r0-r1) = 8.99e-09(2.80e-09C*8.53e-09C)/(1.44e-06m - 5.0e-07m) = 0.23 J

This answer continues to come back wrong. I am supposed to be getting 0.28 J, but I'm not sure why.

The second part should be easy if I get the PE correct.

KE = PE = v(sqrt(m/2)) --> 0.28J/(sqrt(8.6e-09kg/2)) = 4269 m/s

...but that is wrong too. It should be 8070 m/s.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
voko
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391
How did you get the formula for the PE change, and how did you get the formula for speed from KE?
 
  • #3
jbarnhart5
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How did you get the formula for the PE change, and how did you get the formula for speed from KE?

I thought that's what they were. I don't have the book so I'm trying to work from memory. Apparently its not right?
 
  • #4
voko
6,054
391
None of them. The stuff you have in relevant equations, however, is good. You should be able to use that. Especially the PE change should be very straightforward.
 
  • #5
jbarnhart5
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OK, so I think I have this figured out. Using the formula for PE, I calculated with d=1.44e-6 m for PE1 and 5.0e-7 for PE2. Then take PE2 - PE1 and I got the right answer. Finally...thanks for the nudge in the right direction.
Then for speed it was v = sqrt(2KE/M)
 
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