Electrostatic force problem help with vector parts

  • Thread starter Rockdog
  • Start date
  • #1
23
0
I've included a picture.

Two charges Qc and -Qc(Qc = 4 µC) are fixed on the x-axis at x = -7 cm and x = 7 cm, respectively. A third charge Qb = 5 µC is fixed at the origin.
A particle with charge q = 0.3 µC and mass m = 5 g is placed on the y-axis at y = 14 cm and released. There is no gravity.

a) Calculate x-component of acceleration of particle.
b) Calculate y-component of acceleratin of particle.
c) Magnitude of the net electric force on q at its point of release?
=====
Ok, I know the general idea of what I got to do. To get the acceleration, set the electrostatic force equal to mass times acceleration
F=ma
f/m=a

Now I got three charges in a line. For simplicity sake, lets call the charges from left to right in the line 1, 2,3, and charge 4 on the y axis.

Before I go any further, I realize that this is an iscoceles triangle.

When I draw the FBD on charge 4, I have a F4_2(force on 4 by 2) going vertical, F4_1 up to the right, and F4_3 down to the right.

With the use of Coulomb's law, I figured out the electrostatic force from each charge onto charge 4.

|F4_1|= |(k*q_1*q_4)/(.157m)^2| =>.4408N
|F4_2|= |(k*q_2*q_4)/(.14m)^2| => .689N
|F4_3|= |(k*q_3*q_4)/(.157m)^2| => .4408N

.157m is the sides of the triangles
.14m is the height of triangle
q1=4E-6 C
q2=5E-6 C
q3=-4E-6 C
q4=.3E-6 C
=======
Ok, I believe I'm doing good so far.

Now this is where I'm having trouble with. To get the x and y components of acceleration, I have to break up the electrostatic forces...btw, I'm getting these angles by putting the origin on q4 on my FBD.

x-component F4_1
.4408N * cos 70
y-component F4_1
.4408 sin 70

x-component F4_2
0
y-component F4_2
.689N*sin 90

x-component F4_3
.4408N * cos -70
y-component F4_3
.4408 sin -70
----------
Add all the x's together, all the y's together.
Fx=.302 N
Fy=.689 N

a_x=.302N/.005kg=60.4m/sec^2
a_y=.689N/.005kg= 137.8 m/sec^2

Surprisingly, my a_y answer is correct, but my a_x answer is not, which makes me think that I screwed up the x-component calcualtions somewhere, but I do not know where...Sorry for long post, but can't some one help me out with my frustration?
 

Answers and Replies

  • #2
39
0
Ok, I know whats up...just because its iscoceles triangle, it doesn't mean you have 70 degree angles. Just do inv tan of .14/.07 to get 63.4 degrees, and then the x-components will work out.
 
  • #3
156
0
I have no idea what rdn meant by that previous post...

I believe the x component of your acceleration will be zero sue to symmetry.
Qc and -Qc are equal and opposite. The force that each one acts on q will be cancelled by the other.

There will only be a y component of acceleration.
 
  • #4
Janitor
Science Advisor
1,099
1
Paul, the symmetry is not there because the charge on the left is positive while the charge on the right is negative. So the positive test particle (if I may call it that) will be accelerated in a direction that will include a nonzero x component, and more specifically, the x component of acceleration will be in the positive x direction.

Rdn is correct that the angles involved can be calculated by arctan(2).
 
  • #5
156
0
Janitor, your right. Sorry about that, I don't know what I was thinking.

As for the arctan thing, I see it now. I should have drawn it out before making dumb statements. That is what I get for reading it quick and making snap judgements. Sorry gang.
 

Related Threads on Electrostatic force problem help with vector parts

Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
826
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
14
Views
945
Top