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What will be the acceleration of an electron when it is at the origin?

  • Thread starter AxM=Fam
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  • #1
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1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)?


2. K=8.988X10^9 N(m^2/c^2)

Fc= K(qs qd/r^2sd)(direction rsd)
I believe this is the electrostatic force formula? s=source d=detector.


3. This is my attempt at the problem.
Fc=(K)(3.00X10^-6C)(2.50X10^-9C)/(2.5625X10^-4 m^2) [.624695i+.780869j]
The 2.50X10^-9C was a negative to begin with but I redistributed the negative to the i and j hat direction to make them positive.

|Fcsd|=.26306N
θ=Tan^-1(.780869/.624695)=51.3degrees

I cannot figure out how to get the acceleration of this electron?
Do I use F=ma
.26306N/(9.109X10^-31 kg) = 2.89X10^29 m/s^2

I am probably using the wrong formula to begin with can I please get some help with this.
 

Answers and Replies

  • #2
SammyS
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1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)?

2. K=8.988X10^9 N(m^2/c^2)

Fc= K(qs qd/r^2sd)(direction rsd)
I believe this is the electrostatic force formula? s=source d=detector.

3. This is my attempt at the problem.
Fc=(K)(3.00X10^-6C)(2.50X10^-9C)/(2.5625X10^-4 m^2) [.624695i+.780869j]
The 2.50X10^-9C was a negative to begin with but I redistributed the negative to the i and j hat direction to make them positive.

|Fcsd|=.26306N
θ=Tan^-1(.780869/.624695)=51.3degrees

I cannot figure out how to get the acceleration of this electron?
Do I use F=ma
.26306N/(9.109X10^-31 kg) = 2.89X10^29 m/s^2

I am probably using the wrong formula to begin with can I please get some help with this.
Hello AxM=Fam . Welcome to PF !

It looks to me like you're finding the force on one of the charges due to the other.

You need to find the force that each charge exerts on the electron, then add those forces vector-wise.

By the Way: 0.624695i + 0.780869j is not a unit vector.. It's magnitude is approximately 1.4 .
 
  • #3
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I think I am getting a better understanding of the problem. I was missing the charge for the electron which was e=-1.602 x 10^-19. I would just have to figure the net force of the two point charges acting on the electron. From there would I just have to divide the net force by the given electron mass?

I am having a hard time understanding the direction of the vector +3.00μC acting on the electron. Is + always attracted to the negative meaning it will go towards the negative or will the negative go up towards the +. This will help me when I begin to add both vectors to get the net force. I know that like signs will repel so the -2.50nC will push the electron up .5cm.

Can I get more help on this (direction of the +3.00μC on electron is confusing me)? Hope I am at least getting the right idea on how this works.
 
  • #4
gneill
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Opposite charges attract, like charges repel. A negative electron will be pulled towards a positive charge and repelled from a negative charge. Thus the direction of any resulting force vector will depend upon the signs of the interacting charges and their relative positions in the coordinate system. It's always a good idea to draw a diagram and sketch in an approximation of the force vectors.
 

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