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
 

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Answers and Replies

  • #2
8
0
Yes your calculation of a_y is correct.

and a_x by my calculation is 395.4605714 m/sec^2

Fx = F1x + F3x
= F1*cos(x) + F3*cos(x)
where x is angle between x-axis and F1, which is equal to that between x-axis and F3.
and cos(x) = 0.157/0.070 because it's right triangle.

if you calculate above
Fx = 2*F1*cos(x) ----------- because |F1| = |F3|
= 2*F1*(0.157/0.070)
= 1.977302857 N
and a_x = 1.977302857/0.005 = 395.4605717 m/sec^2
 
  • #3
8
0
And by the way, don't forget to tell me whether my calculation is right.

Good luck. :biggrin:
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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964
"x-component F4_1
.4408N * cos 70
y-component F4_1
.4408 sin 70"

Where did you get "70"? Since your right triangles have legs of length 7 and 14, the angle at the base (at 1 and 3) are arctan(14/7)= 63.4 degrees. The opposite angle is 90- 63.4= 26.6 degrees (and the angle at the top of the isosceles triangle is twice that: 53.2 degrees.
The x-component of F4_1 is 0.4408* cos (63.4) and the y-component is
0.4408* sin(64.4)
 

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