Magnitute and Direction of Electric Field Based on Particle

  • #1

Homework Statement


A positively charged particle initially at rest on the ground accelerates upward to 180 m/s in 2.70 s .The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.

What are the magnitude and direction of the electric field?

Express your answer to two significant digits and include the appropriate units. Enter positive value if the electric field is upward and negative value if the electric field is downward.

Homework Equations


What I consider to be the relevant equations:
F=ma
E=F/q
E=ma/q

The Attempt at a Solution


The acceleration, a, is (180m/s)/2.70s=66.67m/s^2

Plugging that into E=ma/q, we get E=m(66.67m/s^2)/q. Since the charge-to-mass ratio is 0.100 C/kg, I figure we can use the reciprocal, 10 kg/C, and use that as m/q.

Which means we get (10kg/C)(66.67m/s^2) = 666.7 N/C.

As for the direction of E, the particle is positively charged and accelerating upward, so it's clear there is a positive field propelling the + charged particle upward. Since electric field always goes from positive to negative, the field is pointed upward, so E is positive.

Rounding for two significant digits, we get 670 N/C.

The only problem is that this answer is wrong. I've also tried 667 N/C and 666.67 N/C just for kicks, and those were also, predictably, wrong. It seems like this should be a really easy problem, which is why it's so frustrating. What am I missing? Am I totally off-base?

Thanks for your help and time,

Matt
 

Answers and Replies

  • #3
Gravity! Of course!

So now 67m/s2 = aE -g => aE = 76 m/s2

Which gives us E=(m/q)(76m/s2)=(10kg/C)(76m/s2) = 760 N/C

Thank you very much, rude man!
 
  • #4
rude man
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Gravity! Of course!

So now 67m/s2 = aE -g => aE = 76 m/s2

Which gives us E=(m/q)(76m/s2)=(10kg/C)(76m/s2) = 760 N/C

Thank you very much, rude man!
Big 10-4, odd guy!
 

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