# Finding the electric field magnitude, and angle below the horizontal

by jheld
Tags: angle, electric, field, horizontal, magnitude
P: 81
1. The problem statement, all variables and given/known data
In the figure below d = 2.6 cm. What is the electric field at the position indicated by the dot in the figure?

2. Relevant equations

E = charge/4*pi*epsilon_0*distance^2
The figure is on the attachment.
3. 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
Attached Files
 field from three point charges.doc (28.5 KB, 28 views)
Mentor
P: 40,240
 Quote by jheld *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?

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