# Electrostatic force problem help with vector parts

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?
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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
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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
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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

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.

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.

Janitor
Science Advisor
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).

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.