Magnitute and Direction of Electric Field Based on Particle

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SUMMARY

The magnitude and direction of the electric field acting on a positively charged particle accelerating upward to 180 m/s in 2.70 seconds is determined to be 760 N/C upward. The charge-to-mass ratio of the particle is 0.100 C/kg, which was used in conjunction with the acceleration due to gravity to calculate the net acceleration. The relevant equations utilized include F=ma and E=ma/q, with the final electric field value rounded to two significant digits as 760 N/C.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of electric field calculations (E=F/q)
  • Familiarity with charge-to-mass ratio concepts
  • Basic principles of gravitational force on charged particles
NEXT STEPS
  • Study the implications of charge-to-mass ratios in electric fields
  • Learn about the effects of gravitational force on charged particles
  • Explore advanced electric field calculations in non-uniform fields
  • Investigate the relationship between acceleration and electric fields in different scenarios
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and kinematics, as well as anyone involved in solving problems related to electric fields and forces on charged particles.

Slightly Odd Guy
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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
 
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Slightly Odd Guy said:
Plugging that into E=ma/q,
There are two forces acting on the particle, not just one!
 
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!
 
Slightly Odd Guy said:
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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