Electric Potential slowing a charged particle to a stop

In summary, the conversation discusses a problem involving a particle with a known mass and charge moving along the positive x-axis with a given speed. It then enters a region of uniform electric field and moves a certain distance before coming to rest. The question asks for the magnitude of the electric field, with multiple answer choices given in kilonewtons per coulomb. However, there is a discrepancy in the question as the charge is given in microcoulombs, which may be a mistake. The person asking the question also mentions having difficulty with unit conversions.
  • #1
fight_club_alum
63
1

Homework Statement


A particle (mass 6.7 × 10–27 kg, charge 3.2 × 10–19 μC) moves along the positive x-axis with
a speed of 4.8 × 105 m/s. It enters a region of uniform electric field parallel to its motion
and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric
field?
a) 2.5 KN/C
b) 1.5 KN/C
c) 1.2 KN/C
d) 3.5 KN/C
e) 2.4 KN/C

Homework Equations


(delta)E.E = -q * (E.d)

The Attempt at a Solution


(1/2*(6.7*10^-27)*(4.8*10^5)^2)/( (3.2*10^-19) * 10^-6) *2)

The problem is that when I convert the charge from uc to c I get the answer but multiplied by a number like million and when I divide it by 1000 to get KN still the number remains multiplied by a factor of 1/10. On the other hand, if I leave the charge in uc I get the answer when I divide by 1000 as I should. I feel that I messed a step in the middle or there is a mistake in the question
Thanks in advance
 
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  • #2
fight_club_alum said:

Homework Statement


A particle (mass 6.7 × 10–27 kg, charge 3.2 × 10–19 μC) moves along the positive x-axis with
a speed of 4.8 × 105 m/s. It enters a region of uniform electric field parallel to its motion
and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric
field?
a) 2.5 KN/C
b) 1.5 KN/C
c) 1.2 KN/C
d) 3.5 KN/C
e) 2.4 KN/C

Homework Equations


(delta)E.E = -q * (E.d)

The Attempt at a Solution


(1/2*(6.7*10^-27)*(4.8*10^5)^2)/( (3.2*10^-19) * 10^-6) *2)

The problem is that when I convert the charge from uc to c I get the answer but multiplied by a number like million and when I divide it by 1000 to get KN still the number remains multiplied by a factor of 1/10. On the other hand, if I leave the charge in uc I get the answer when I divide by 1000 as I should. I feel that I messed a step in the middle or there is a mistake in the question
Thanks in advance
The smallest mass a particle (electron or proton) can have is 1.6 *10-19C. An ionized particle can have a multiple of it. So the μC should be a mistake.
 
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Likes fight_club_alum
  • #3
ehild said:
The smallest mass a particle (electron or proton) can have is 1.6 *10-19C. An ionized particle can have a multiple of it. So the μC should be a mistake.
yes, it turned out that the question is written incorrectly; (uc) should be (c) in this question.
Thank you so much
 
  • #4
Think about the kinematic equations. What's the acceleration? Then use ##F=ma##
 
  • #5
Miles123K said:
Think about the kinematic equations. What's the acceleration? Then use ##F=ma##
That was not the issue. See posts 2 and 3.
 

Related to Electric Potential slowing a charged particle to a stop

1. What is electric potential?

Electric potential is a measure of the amount of electric potential energy per unit charge at a certain point in an electric field.

2. How does electric potential slow a charged particle to a stop?

When a charged particle moves through an electric field, it experiences a force that is opposite in direction to the direction of its motion. This force acts to slow down the particle, eventually bringing it to a stop.

3. How does the strength of the electric field affect the slowing of a charged particle?

The strength of the electric field directly affects the force acting on the charged particle. A stronger electric field will result in a greater force, causing the particle to slow down more quickly.

4. Can electric potential be used to accelerate a charged particle?

Yes, electric potential can be used to accelerate a charged particle. This is achieved by applying an electric potential difference between two points, creating an electric field that causes the particle to gain speed.

5. Is there a limit to how much electric potential can slow a charged particle?

Yes, there is a limit to how much electric potential can slow a charged particle. This limit is reached when the particle has come to a complete stop and is at the same electric potential as its surroundings. At this point, there is no longer a force acting on the particle to slow it down further.

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