We will see that the electric field in that point is:
Ex = 31.598 N/μc
Remember that for a charge Q, the electric field is given by:
[tex]E = k*\frac{Q}{r^2} (n_x, n_y)[/tex]
Where the versor is in the direction of r, k is a constant, and r is the distance to the charge.
In this case, we have two charges, one at (0cm, 5cm) and the other at (0cm, -5cm). Because of the symmetry on the y-axis, we know that the y-component will be canceled, so we only have the x-component.
[tex]E_x = k*(\frac{30 \mu C}{(12cm)^2 + (5cm)^2} + \frac{30 \mu C}{(12cm)^2 + (-5cm)^2} )\\\\E_x = k*(\frac{60 \mu C}{169cm^2})[/tex]
Now we know that the value of k, the Coulomb's constant is:
k = 8,9*10^9 N*m^2/C^2
Writing this in cm and μc (multiplying by 10^4 and dividing by 10^12 we get)
k = 8.9*10^9*(10^4*10^-12) N*cm^2/μc^2
k = 8.9*10 N*cm^2/μc^2
Now we use that in our equation:
[tex]E_x = (8.9*10 N*cm^2/\mu c^2)*(\frac{60 \mu C}{169cm^2}) = 31.598N/\mu c[/tex]
This is the electric field on the point (12cm, 0cm).
If you want to learn more about electric fields, you can read:
https://brainly.com/question/1592046
