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Finding 2 unknow charges

  • Thread starter bobred
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  • #1
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Homework Statement


There are two opposite charges of equal magnetude, Q1 and Q2 separated by 4m. There is a central segment of 0.1m AB located centrally between the two charges AB. The electric field between AB can be taken as uniform. An electron is released with negligable speed at A and passes B 0.015 s later. Find the magnitude and sign of both Q1 and Q2.


Homework Equations


[tex]s=ut+\frac{1}{2}at^{2}[/tex]


[tex]F_{el}=am_{e}[/tex] so [tex]a=\frac{F_{el}}{m_{e}}=\frac{q\mathcal{E}}{m_{e}}=\frac{-e\mathcal{E}}{m_{e}}[/tex]


[tex]F_{el}=k\frac{\left|q\right|\left|Q\right|}{r^{2}}[/tex] where [tex]k=\frac{1}{4\pi\epsilon_{0}}=8.988\times10^{9}\,\textrm{N}\,\textrm{m}^{2}\,\textrm{C}^{-2}[/tex]


[tex]\mathcal{E}=\frac{m_{e}a}{-e}[/tex]


[tex]m_{e}=9.109\times10^{-31}\,\textrm{kg}[/tex]


[tex]e=-1.602\times10^{-19}\,\textrm{C}[/tex]


The Attempt at a Solution


Taking the positive x direction as the direction of the electron,
we can find the acceleration by

[tex]a=\frac{2s}{t^{2}}=\frac{2\times0.1\,\textrm{m}}{(0.015\,\textrm{s})^{2}}=888.89\,\textrm{m}\,\textrm{s}^{2}[/tex]


The electric field is


[tex]\mathcal{E}=\frac{am_{e}}{e}=\frac{888.89\,\textrm{m}\,\textrm{s}^{2}\times9.109\times10^{-31}\,\textrm{kg}}{-1.602\times10^{-19}\,\textrm{C}}=-5.054\times10^{-9}\,\textrm{N}\,\textrm{C}^{-1}[/tex]


The distance from Q2 to be is 1.95 m so the force at B is


[tex]F_{el}=q\mathcal{E}=-1.602\times10^{-19}\,\textrm{C}\times-5.054\times10^{-9}\,\textrm{N}\,\textrm{C}^{-1}=8.097\times10^{-28}\,\textrm{N}[/tex]


Using Coulumb's law and rearranging we find the charge


[tex]\left|Q\right|=\frac{F_{el}r^{2}}{k\left|q\right|}=\frac{8.097\times10^{-28}\,\textrm{N}\times(1.95\,\textrm{m})^{2}}{8.988\times10^{9}\,\textrm{N}\,\textrm{m}^{2}\,\textrm{C}^{-2}\times1.602\times10^{-19}\,\textrm{C}}=2.138\times10^{-18}\,\textrm{C}[/tex]

As the electron is travelling toward B we can tell that Q2 is +ve.

Does what I have done look ok?
 
Last edited:

Answers and Replies

  • #2
gneill
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20,793
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Why not take the midpoint between your charges as the place where the field is the average value, and find |Q| from that?
 
  • #3
173
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Good point

[tex]
\left|Q\right|=\frac{F_{el}r^{2}}{k\left|q\right|} =\frac{8.097\times10^{-28}\,\textrm{N}\times(2.00\,\textrm{m})^{2}}{8.988 \times10^{9}\,\textrm{N}\,\textrm{m}^{2}\,\textrm{ C}^{-2}\times1.602\times10^{-19}\,\textrm{C}}=2.249\times10^{-18}\,\textrm{C}
[/tex]

What I wanted to know was if my method was sound.
 
  • #4
gneill
Mentor
20,793
2,773
Remember that there are two charges Q involved. Each will produce a force of magnitude Fel at the center point. Otherwise, your method looks good.
 

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