# Magnitute and Direction of Electric Field Based on Particle

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1. Apr 15, 2016

### Slightly Odd Guy

1. The problem statement, all variables and given/known data
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.

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

3. 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

2. Apr 15, 2016

### rude man

There are two forces acting on the particle, not just one!

3. Apr 15, 2016

### Slightly Odd Guy

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. Apr 15, 2016

### rude man

Big 10-4, odd guy!

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