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KE and speed of a sphere near a fixed charge

  • Thread starter jbarnhart5
  • Start date
  • #1

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
6,054
390
How did you get the formula for the PE change, and how did you get the formula for speed from KE?
 
  • #3
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
6,054
390
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
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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