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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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# Homework Help: Electrostatic force problem help with vector parts

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