Electric field problem

[Punchlinegirl]

An equilateral triangle with sides 66 m has charges +5 $$\mu C$$, -5 $$\mu C$$, and -9.3 $$\mu C$$ at its vertices and charge + 18.6 $$\mu C$$ at the midpoint on its side between the vertices with charges $$+5 and -5 \mu C$$. What is the magnitude of the x-component of the electric field at point P, the point forming a second equilateral triangle with +5 $$\mu C and -5 \mu C$$? Answer in units of N/C.
Sorry I don't have a picture but it's 2 equilateral triangles that look like a mirror image of each other.
First I drew a picture of what the electric fields would look like. The -9.3 and 18.6 $$\mu C$$ would not have a y-component, so I ignored those for now.
so I found that
k * 5 x 10^-6/ 66^2 * 33/66 = 5.17
and k * -5 x 10^-6 / 66^2 *57.2/66 = 8.95 (since it's absolute value of the charge)
So E_x= 14.12... which isn't right.. help please?

 

[Punchlinegirl]

Ok I figured out what I was doing wrong... the second part of the problem asks for the magnitude of the y-component of the electric field at point P.
E_1 = k * 5 x 10^-6 / 66^2 * 57.2/66 = 8.93
E_2 = 8.93 (same set up as E_1)
E_3 = k * 18.6 x 10^-6 / 57.2^2 = -51.2, since the field would be downward
E_4 = k * 9.3 x 10^-6 / 114.4^2 = 6.40
8.93+8.93-51.2+6.40= -26.94
This isn't right.. can someone help me?

 

[kyle8921]

I would approach this problem by first identifying the key information and variables given. The equilateral triangle has sides of 66 m, with charges of +5, -5, and -9.3 \mu C at the vertices, and +18.6 \mu C at the midpoint between the vertices with +5 and -5 \mu C. We are also asked to find the magnitude of the x-component of the electric field at point P, which forms a second equilateral triangle with +5 \mu C and -5 \mu C.

To solve this problem, we can use the equation for electric field strength, E = kQ/r^2, where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge.

First, we can calculate the electric field at point P due to the charges at the vertices of the first equilateral triangle. Using the equation above, we get:

E1 = k(5 x 10^-6 C)/ (66 m)^2 = 5.17 N/C

Similarly, for the second equilateral triangle, we can calculate the electric field at point P due to the charges at the vertices:

E2 = k(-5 x 10^-6 C)/ (66 m)^2 = -5.17 N/C

Now, we need to take into account the electric field at point P due to the charge at the midpoint between the +5 \mu C and -5 \mu C charges. To do this, we can use the principle of superposition, which states that the total electric field at a point due to multiple charges is the vector sum of the individual electric fields at that point.

The electric field at point P due to the charge at the midpoint is given by:

E3 = k(18.6 x 10^-6 C)/ (33 m)^2 = 14.12 N/C

Since this charge is located along the x-axis, the y-component of the electric field due to this charge will cancel out.

Now, we can find the total electric field at point P by adding the individual electric fields together:

E_total = E1 + E2 + E3 = 5.17 N/C - 5.17 N/C + 14.12 N/C = 14.12 N/C

Therefore, the magnitude of