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

Does what I have done look ok?

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