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bobred
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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 traveling toward B we can tell that Q2 is +ve.
Does what I have done look ok?
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