# Force exerted by an external electric field

• lacar213
In summary, the problem involves a particle of mass 1.20 kg and a charge of +2.30 mC being shot into an electric field of 18.0 N/C in the opposite direction of the particle's velocity. With an initial speed of 2.40 m/s, the particle comes to a complete stop in approximately 4.08 seconds. This is calculated using the equation Vf = Vi +at, taking into account the force of the electric field (F=qE) and the mass of the particle (F=ma). The final answer is found to be 0.000014 seconds, after correcting for units and arithmetic errors.
lacar213

## Homework Statement

A particle of mass 1.20 kg with a charge of +2.30 mC is shot into a region with an electric field of strength 18.0 N/C, having a direction opposite the particle's velocity. If the particle has an initial speed of 2.40 m/s, how long does it take to come to a complete (though momentary) standstill? Assume there are no other forces acting on the particle.

F = qE
F= ma

## The Attempt at a Solution

I'm not sure what equation to start with because of all the information already given. I couldn't find any equations dealing with time in the chapter this problem was in. It sounds like you may have to use kinematics but I'm not sure if that is the correct way to do it.

Kinematics (well dynamics really) sounds good to me. What do you think the first step would be?

Vf = Vi +at ??

lacar213 said:
Vf = Vi +at ??
Looks good to me.

0 = 2.40 + 9.8 (t)
t = 9.8 / 2.4 = 4.08333

lacar213 said:
0 = 2.40 + 9.8 (t)
t = 9.8 / 2.4 = 4.08333
Why have you used an acceleration of 9.8 m.s-2?

I forgot to find the force using F=qe then you can plug it into f=ma

A = .000035

.000035t = 2.40 - 0
t = .000014

would the acceleration be negative also since the field of strength is opposite the particle

the problem states millicoulombs not micro - t = .014375

lacar213 said:
I forgot to find the force using F=qe then you can plug it into f=ma

A = .000035

.000035t = 2.40 - 0
You're good up until this point. However, your final answer is wrong.
lacar213 said:
t = .000014
You may want to check your arithmetric
lacar213 said:
would the acceleration be negative also since the field of strength is opposite the particle
Yes, but you took care of that (be it intentionally or otherwise) in your second line above, by multiplying through by -1.

lacar213 said:
the problem states millicoulombs not micro - t = .014375
Your final answer is still wrong. Recheck your basic arithmetic. Notice that you are dividing a number greater than one by a number smaller than one. How can the result be less than one?

2.4 / .0345 = 69.5652

lacar213 said:
2.4 / .0345 = 69.5652
You're getting closer, but your acceleration is two orders of magnitude out.

GOT IT thanks for the help!

lacar213 said:
GOT IT thanks for the help!
Not a problem

## 1. What is the definition of force exerted by an external electric field?

The force exerted by an external electric field is the force that a charged particle experiences when placed in an electric field. It is a vector quantity that is dependent on the magnitude and direction of the electric field as well as the charge of the particle.

## 2. How is the force exerted by an external electric field calculated?

The force exerted by an external electric field can be calculated using the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength. This equation is based on Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

## 3. What are the units of force exerted by an external electric field?

The units of force exerted by an external electric field are newtons (N) in the SI system and dyne (dyn) in the CGS system. Both units are equivalent and represent the amount of force required to accelerate a mass of one kilogram by one meter per second squared.

## 4. How does the direction of the force exerted by an external electric field depend on the charge of the particle?

The direction of the force exerted by an external electric field on a charged particle depends on the sign of the charge. If the charge is positive, the force will be in the direction of the electric field, while if the charge is negative, the force will be in the opposite direction of the electric field.

## 5. Can the force exerted by an external electric field change the kinetic energy of a charged particle?

Yes, the force exerted by an external electric field can change the kinetic energy of a charged particle. If the force is in the same direction as the particle's velocity, it will increase the kinetic energy, while if the force is in the opposite direction, it will decrease the kinetic energy. This is because the work done by the force is equal to the change in kinetic energy of the particle.

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