KE and speed of a sphere near a fixed charge

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Homework Help Overview

The problem involves a tiny sphere with a specific mass and charge, released from rest near a fixed charge. The tasks are to find the kinetic energy and speed of the sphere at a given distance from the fixed charge, utilizing principles of potential and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, specifically the relationship between potential energy (PE) and kinetic energy (KE). There are questions about the derivation of formulas used for calculating changes in potential energy and speed from kinetic energy.

Discussion Status

Some participants have attempted to clarify their understanding of the equations involved, while others have provided guidance on the correct approach to calculating potential energy changes. There is acknowledgment of errors in initial calculations, and some participants express confidence in their revised methods.

Contextual Notes

Participants mention a lack of access to reference materials, which may affect their ability to verify formulas and concepts. There is also a noted discrepancy between expected and calculated values for kinetic energy and speed.

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.
 
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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?
 
voko said:
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?
 
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.
 
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)
 
Last edited:

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