Electric Fields , Finding final velocity. NEED HELP

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Homework Help Overview

The problem involves calculating the final velocity of a charged mass (Q1) as it moves in the electric field created by another charge (Q2). The context includes concepts from electrostatics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of various formulas, noting that the original poster's use of kinematic equations may not be appropriate due to changing acceleration. Suggestions are made to consider energy conservation instead.

Discussion Status

Some participants have provided guidance on using the work-energy principle, indicating a shift in approach from kinematics to energy methods. There is an acknowledgment of the original poster's calculations and the discrepancy with the answer key, but no consensus has been reached on the final outcome.

Contextual Notes

Participants highlight the importance of recognizing that acceleration is not constant in this scenario, which affects the choice of equations used. The original poster's calculations and the answer key's result are noted as points of discussion.

skye204
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Homework Statement


Mass= 15g
Q1= 8.0uC
Q2= 25 uC (stationary)

Q1 is released from a point 12cm from Q2. How fast will Q1 be moving when the separation is 20cm.


Homework Equations



F=ma
F=qe
F=kQq/r2
E=kQ/r2
Ke=mv2/2
W=FD
vf2=vi2+2ad

The Attempt at a Solution


F=(9x109)(25x10^-6)(8x10^-6)/(.12^2)
F=125N

F=MA
125/0.015= A
A= 8333m/s2

vf2=vi2+2ad
vf2=0+2(8333)(.20-.12)
vf=36.5m/s

ACTUAL ANSWER IS : 28.3 according to the answer key

Thanks for help!
 
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Hi skye204! Welcome to PF :smile:

The formula that you've used i.e "vf2=vi2+2ad" is applicable only when the acceleration is constant. In this case acceleration changes, so this is not applicable. I would suggest you apply the energy conservation principle instead :wink:
 
vf2=vi2+2ad
vf2=0+2(8333)(.20-.12)
vf=36.5m/s



you can not use the above formula as the acceleration is not constant.
you should try using energy method.
 
I get it

W=FD
W=125N x.12
W= 15

F=(9x109)(25x10^-6)(8x10^-6)/(.2^2)
F=45N


W=45 x .2
W= 9

pe=pe' + ke'
15= 9 + 0.015(v^2)/2
v= 28.28

Thanks!
 

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