Two point particles, each having a charge equal to q, sit on the base of an equi

Question

Two point particles, each having a charge equal to q, sit on the base of an equilateral triangle that has sides of length L.

Basically there are 3 point particles, each occupy a respective vertex of an equilateral triangle. 2 of the vertex point particles at the base of the triangle have a charge of +q; the third point particle sitting at the apex has a charge of +2q.

A fourth point particle that has charge q' is placed at the midpoint of the baseline making the electric field at the center of the triangle equal to zero. What is the value of q'? (The center is in the plane of the triangle and equidistant from all three vertices.)

I got q' = q/4 as an answer but I am not sure if that is correct.

Summing the electric fields in the x-direction results in 0.

Summing in the y-direction is what I believe I did incorrectly:

0 = kqsin30 + kq/d^2 - 2kq/d^2 + kq'/(dsin30)^2 ; where d is the distance from the vertex to the center of the equilateral triangle. Multiplying by d^2/k and rearranging left me with: q' = q/4
If I did make a mistake, please explain how you arrived at your set-up. There is an answer posted to a very similar question but the y-direction sum set-up does not make much sense to me. Thanks. Two point particles, each having a charge equal to q, sit on the base of an equilateral triangle that has sides of length L.

Basically there are 3 point particles, each occupy a respective vertex of an equilateral triangle. 2 of the vertex point particles at the base of the triangle have a charge of +q; the third point particle sitting at the apex has a charge of +2q.

A fourth point particle that has charge q' is placed at the midpoint of the baseline making the electric field at the center of the triangle equal to zero. What is the value of q'? (The center is in the plane of the triangle and equidistant from all three vertices.)

I got q' = q/4 as an answer but I am not sure if that is correct.

Summing the electric fields in the x-direction results in 0.

Summing in the y-direction is what I believe I did incorrectly:

0 = kqsin30 + kq/d^2 - 2kq/d^2 + kq'/(dsin30)^2 ; where d is the distance from the vertex to the center of the equilateral triangle. Multiplying by d^2/k and rearranging left me with: q' = q/4
If I did make a mistake, please explain how you arrived at your set-up. There is an answer posted to a very similar question but the y-direction sum set-up does not make much sense to me. Thanks. 0 = kqsin30 + kq/d^2 - 2kq/d^2 + kq'/(dsin30)^2 ; where d is the distance from the vertex to the center of the equilateral triangle. Multiplying by d^2/k and rearranging left me with: q' = q/4

Explanation / Answer

Let me know if you want me to write the solution before rating. x-set up looks like 0 = kq/(dcos30)^2 - kq/(dcos30)^2 which is true y-set up looks like 0 = kqsin30/d^2 + kqsin30/d^2 + kq'/(dsin30)^2 - 2kq/d^2
=>q' = q/4

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