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# Charge problems

QUESTION GIVEN:

1) In the figure, a central particle of charge -q is surrounded by two circular rings of charged particles. What are the magnitude and direction of the net electrostatic force on the central particle due to the other particles. (Hint: Symmetry can reduce work)

In the figure there is a circle, then another circle inside. Now, draw a y axis and an x axis so the origin is in the center of the inner circle or in other words both circles are quartered now. Simple. Now placed two more lines so the circles are divided into eights. Like a pie inside a pie all cut the same way.

Now, the outside circle has a charge at the top of the y axis on the circle edge of +4q and at the bottom of the y axis on the circle edge +4q. The outmost edges of the x axis of the circle have no charges. NEXT, when you cut the circles into eights... place a -2 q charge going along the circle edge to the left of the top y axis. The charge is - 2q, then -2q to the right along the edge of the circle where you placed the eight dividing lines. The bottom of the y axis was +4q, now place a -2q to the left following the circle edge to the 1/8th disecting line and to the right -2q. SUMMARY: starting at the top of the out circle was +4q, going the the right following the edge and stopping at each disecting line to place a charge is - 2q, then no charge, then -2q, then at the bottom +4q, - 2q, now at the left side of the circle no charge, then -2q and back to the top again +4q.

The inner circle at the top is +2q going to the right, +q, then -7q, then -2q, then no charge, +q, then -7q, no chage and to the top again +2q.

CONFUSING, but I know if two charges are the same and opposite they cancel. So, all charges cancel except for the charge at the top of the inner circle which is +2q.

2) Figure shows 4 arrangements of charged particles. Rank the arrangements according to the magnitude of the net electrostatic force on the particle with charge +Q greatest first.

Arrangement 1:
-x axis and a +y axis:
+Q going on left of -x axis then charge p at origin, then charge p at the top of + y axis. distance of -x axis is 2d and distance of + y axis is d.

Arrangement 2:

-x axis and a +y axis:
+Q going on left of -x axis then charge e at origin, then charge p at the top of + y axis. distance of -x axis is 2d and distance of + y axis is d.

Arrangement 3:
-x axis and a +y axis:
+Q going on left of -x axis then charge p at origin, then charge e at the top of + y axis. distance of -x axis is 2d and distance of + y axis is d.

Arrangement 4:
-x axis and a +y axis:
+Q going on left of -x axis then charge e at origin, then charge e at the top of + y axis. distance of -x axis is 2d and distance of + y axis is d.

#### Solution Preview

Ok, F in magnitude is proportional to q1q2/(D)^2 so my line can be along the i unit vector.
I can also set the charge q3 to be -e q where q is a positive number
Same charges repel and opposite charges attract then for line 1 that charges to the left repel q3 then the ...

#### Solution Summary

This solution provides a detailed description on how to find the direction and magnitude net electrostatic force.

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