Finding the electric field magnitude, and angle below the horizontal

[jheld]

Homework Statement

In the figure below d = 2.6 cm. What is the electric field at the position indicated by the dot in the figure?

Homework Equations

E = charge/4*pi*epsilon_0*distance^2
The figure is on the attachment.

The Attempt at a Solution

I decided to first find only the magnitude from each of the fields on the point.
charge 1 = 10 nC;
charge 2 = 5 nC;
charge 3 = -5nC
*I put all of the charges into Coulomb before any calculations
E1 = Kc1/5d^2;
E2 = Kc2/d^2;
E3 = kc3/4d^2;
After adding all of the Es together I get the wrong answer (as far as magnitude).
Answer is: 7.9 × 104 N/C, at 5.2° below the horizontal

 

[Doc Al]

*I put all of the charges into Coulomb before any calculations
E1 = Kc1/5d^2;
E2 = Kc2/d^2;
E3 = kc3/4d^2;
After adding all of the Es together I get the wrong answer (as far as magnitude).

You can't just add up the magnitudes like ordinary numbers (if that's what you did). You must add them as vectors. What is the direction of each field?

 

[nlightner]

I would like to point out that the attempted solution has some errors and may not provide an accurate answer. First, the equation used for electric field magnitude is incorrect. The correct equation is E = k*q/r^2, where k is the Coulomb constant, q is the charge, and r is the distance. Also, the charges given in the problem are in nanoCoulombs (nC), but the attempted solution uses Coulombs (C). It is important to use the correct units in calculations to avoid errors.

Additionally, the attempted solution does not take into account the direction of the electric fields. Each of the charges will have a different direction of electric field and these should be considered when adding them together. This can be done using vector addition.

To find the electric field at the position indicated by the dot, we can use the equation E = k*Σq/r^2, where Σq is the sum of all the charges and r is the distance from each charge to the point. We also need to consider the direction of the electric fields by using vector addition.

Therefore, the correct solution would be to first convert the charges to Coulombs (10 nC = 1e-8 C, 5 nC = 5e-9 C, -5 nC = -5e-9 C). Then, we can use the equation E = (9*10^9)*(1e-8+5e-9-5e-9)/(0.026)^2 = 2.0*10^6 N/C. To find the angle below the horizontal, we can use the inverse tangent function, giving us an angle of 5.2° below the horizontal.

In conclusion, it is important to use the correct equations and units when solving problems in physics. Additionally, considering the direction of vectors is crucial in accurately determining the electric field at a certain point.