Magnitute and Direction of Electric Field Based on Particle

In summary, the conversation discussed the problem of finding the magnitude and direction of the electric field for a positively charged particle accelerating upward on the ground. The relevant equations used were F=ma, E=F/q, and E=ma/q. After some incorrect attempts, it was realized that the force of gravity also needed to be taken into account. With this adjustment, the correct answer of 760 N/C was found.
  • #1
Slightly Odd Guy
11
0

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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  • #2
Slightly Odd Guy said:
Plugging that into E=ma/q,
There are two forces acting on the particle, not just one!
 
  • #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
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!
 

FAQ: Magnitute and Direction of Electric Field Based on Particle

1. What is the concept of electric field based on particle?

The concept of electric field based on particle refers to the force field that surrounds a charged particle, exerting a force on other charged particles within its proximity. The strength and direction of this field are determined by the magnitude and sign of the charged particle.

2. How is the magnitude of an electric field calculated?

The magnitude of an electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. This can be represented by the equation E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What is the unit of electric field magnitude?

The unit of electric field magnitude is newtons per coulomb (N/C) in the SI (International System of Units) system. In the CGS (centimeter-gram-second) system, the unit is expressed as dynes per unit charge (D/cm).

4. How does the direction of the electric field relate to the particle's charge?

The direction of the electric field is always away from a positively charged particle and towards a negatively charged particle. This is because like charges repel each other and opposite charges attract, resulting in an outward or inward direction of the electric field lines, respectively.

5. Can the direction of the electric field change?

Yes, the direction of the electric field can change depending on the location and movement of charged particles. It can also be influenced by the presence of other charged particles or external electric fields.

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