Electric field problem with acceleration

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BoldKnight399
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An electron moving through an electric
field experiences an acceleration of 7200 ×
10^3 m/s2.
Find the magnitude of the electric force
acting on the electron. The Coulomb constant
is 8.99 × 109 N · m2/C2 and the fundamental
charge is 1.60 × 10−19

So I thought that since E*q=F then it would also be the same as F=m*a so it would be:
E*q=m*a
(E)*(1.60e-19 C)=(9.11e-31kg)(7200e3 m/s^2)
which when you do out the calculations would get you E=4.0995e-5.
Somehow I think Newton is crying in the grave with this witchcraft I just created.
Did I have the right idea or am I way off?
 
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looks good to me
 
BoldKnight399 said:
An electron moving through an electric
field experiences an acceleration of 7200 ×
10^3 m/s2.
Find the magnitude of the electric force
acting on the electron. The Coulomb constant
is 8.99 × 109 N · m2/C2 and the fundamental
charge is 1.60 × 10−19

So I thought that since E*q=F then it would also be the same as F=m*a so it would be:
E*q=m*a
(E)*(1.60e-19 C)=(9.11e-31kg)(7200e3 m/s^2)
which when you do out the calculations would get you E=4.0995e-5.
Somehow I think Newton is crying in the grave with this witchcraft I just created.
Did I have the right idea or am I way off?

I haven't thought through this method myself; but I assume that from what the question gives you the approach should be slightly different. I think this because I don't see the reason for giving you the Coulomb constant, unless it is a Red-Herring and I fell for it=]
I'm not sure tbh, Electric fields are not my strong point.
 
I tried it and it was wrong. So I guess that I did something wrong. Any ideas?
 
BoldKnight399 said:
[...]
Find the magnitude of the electric force
acting on the electron.
[...]
(E)*(1.60e-19 C)=(9.11e-31kg)(7200e3 m/s^2)
which when you do out the calculations would get you E=4.0995e-5.
[...]

According to the problem statement, it's asking for the electric force, not the electric field strength. So I guess you only really need F = ma.
 
Moral of the story: I need to learn to read. Thank you so much!