help (1 Viewer)

yashbb · Jan 5, 2022

how do i do part e

Lith_30 · Jan 5, 2022

Use $E=\frac{V}{d}$ to find the strength of the magnetic field.

$E=\frac{200}{0.15}$

Then use $E=\frac{F}{q}$ to find out the force that is acting on the electron

$\\\frac{200}{0.15}=\frac{F}{1.602\times{10^{-19}}}\\\\F=\frac{200\times1.602\times{10^{-19}}}{0.15}$

Now using $F=ma$ we can find the acceleration of the particle.

$\\\frac{200\times1.602\times{10^{-19}}}{0.15}=9.11\times10^{-31}\times{a}\\\\a=\frac{200\times1.602\times{10^{-19}}}{0.15\times9.11\times10^{-31}}$

Then the question basically becomes a projectile motion problem where gravity in this case is the acceleration that was just calculated.

yashbb · Jan 8, 2022

Lith_30 said:
Use $E=\frac{V}{d}$ to find the strength of the magnetic field.

$E=\frac{200}{0.15}$

Then use $E=\frac{F}{q}$ to find out the force that is acting on the electron

$\\\frac{200}{0.15}=\frac{F}{1.602\times{10^{-19}}}\\\\F=\frac{200\times1.602\times{10^{-19}}}{0.15}$

Now using $F=ma$ we can find the acceleration of the particle.

$\\\frac{200\times1.602\times{10^{-19}}}{0.15}=9.11\times10^{-31}\times{a}\\\\a=\frac{200\times1.602\times{10^{-19}}}{0.15\times9.11\times10^{-31}}$

Then the question basically becomes a projectile motion problem where gravity in this case is the acceleration that was just calculated.

Thank you so much

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yashbb

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Lith_30

o_o

yashbb

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