I have worked out this problem and I got. (1/52)q The problem is: Two point part

Question

I have worked out this problem and I got. (1/52)q
The problem is:

Two point particles, each having a charge equal to q, sit on the base of an equilateral triangle that has sides of length L as shown in the figure below. A third point particle that has a charge equal to Q = 13q sits at the apex of the triangle. 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 worked out:
((kqsin30)/d^2)+((kqsin30)/d^2))+(13qi/(dsin30)^2)-(2kq/d^2)=0

I multiplied by d^2/k
and got
.5q+.5q+(13qi/(dsin30)^2) -2q

I kept going and I got qi=(1/51)q

I'm not sure where I messed up but if you could help that would be great!

Explanation / Answer

if d1 = L / square root of(3), and d2 = L /2xsquare root of (3),
then for the net electric field at the centre of triangle is to be zero, we should have the condition
K x 13 q / d1^2 = K x q / d1^2 + K x q' / d2^2
solving this after using the values you can get q' =

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