An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What are the magnitude and direction of the electric field?

Question

anonymous · Answer

$$ \textbf F = m_{e^-} \textbf a  
\ \
\ \textbf F_e = q_{e^-} \textbf E$$

$$ |\textbf E| = \frac{m_{e^-} |\textbf a|}{q_{e^-}} $$
That's the magnitude of the E field, and it points from North to South.

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anonymous · Answer

in ur equation it looks like u are subtracting :D :D!!

anonymous · Answer

Awww man, you're totally right :(  Didn't even think of that!  dumb electrons :P  

^_^

anonymous · Answer

just get rid of the minus :P

anonymous · Answer

$$\textbf F = m_e \textbf a_e$$
$$ \textbf F = q_e \textbf E$$

$$ |\textbf E| = \frac{m_e |\textbf a|}{q_e}$$

already on it ^_^

anonymous · Answer

@Mashy  thanks for catching that - I think the answer might have been more than a little different with minuses there!

anonymous · Answer

haha.. no problem man!.. i just get annoyed to use that whole math thing.. i just dunno how u get the patience to type it out so much :O :O ..

anonymous · Answer

it looks so pretty it tempts me into using it >.>  steals my time and my heart...