Point Charge Electric Field Problem

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SUMMARY

The forum discussion centers on solving an electric field problem involving two point charges of 3.0nC located 10cm apart along the Y-axis. The user attempted to calculate the electric field at a point 5cm to the right of the upper charge using the formula E = [Kq/r^2] * Unit Vector, but arrived at a more complex solution than the book's answer of 1,180 N/C at 5.5 degrees above the X-axis. The discussion highlights the importance of simplifying calculations and following a structured approach to problem-solving in electrostatics.

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  • Understanding of electric fields and point charges
  • Familiarity with the formula E = [Kq/r^2] * Unit Vector
  • Knowledge of trigonometric functions for vector resolution
  • Basic grasp of Pythagorean theorem for vector magnitude calculations
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Rettro
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I've got a test coming up on this, and got quite a few of the questions on the related homework wrong. Attached is the problem, my solutions, and the book(correct) solution, please try to find what I'm doing wrong, this is driving me nuts.


Homework Statement



Two charges exist along the Y axis at Y=0cm and Y=10cm, both at 3.0nC. What is the electric field at (5cm,10cm), 5cm right of the upper charge.

The book has a picture of this, my description may not be perfect. Simply, two 3.0nC charges 10cm apart one above the other, what is the electric field 5cm to the right of the upper charge.

Homework Equations



Point Charge Field, E = [Kq/r^2] * Unit Vector
Pythagorean Theorem, A^2+B^2=C^2

The Attempt at a Solution



E1, lower charge. E2, Upper charge.

E1U --> E1 Unit Vector
E1U --> tan^-1(0.05m/0.1m)
E1U --> 26.565 degrees
E1U --> sin(26.565 degrees) = 0.4472 i
E1U --> cos(26.565 degrees) = 0.8944 j
E1U = 0.4472 i + 0.8944 j
(Check, 0.4472^2 + 0.8944^2 = 0.99999. Unit Vector)

E1 = [Kq/r^2] * Unit Vector
E1 = [(9E9 Nm^2/C^2)(9E-9 C) / 0.0125 m^2] * (0.4472 i + 0.8944 j)
E1 = [(27 Nm^2/C) / (0.0125 m^2)] * (0.4472 i + 0.8944 j)
E1 = 2,160 N/C * (0.4472 i + 0.8944 j)
E1 = 965.952 N/C i + 1,931.904 N/C j

E2 = [Kq/r^2] * Unit Vector
E2 = [(9E9 Nm^2/C^2)(9E-9 C) / (0.05 m)^2] * i (Right(X) Vector)
E2 = [27 Nm^2/C / 0.0025 m^2] * i
E2 = 10,800 N/C i

E = E1 + E2
E = 965.952 N/C i + 1,931.904 N/C j + 10,800 N/C i
E = 11,765.952 N/C i + 1,931.904 N/C j
Emag = sqrt(11,765.952^2 + 1,931.904^2)
Emag = sqrt(142,169,879.5)
Emag = 11,923.5
Edir = tan-1(1,931.904 / 11,765.952) <-- Degrees over X axis
Edir = 9.324 degrees

E = 11,923.5 N/C <9.324 degrees above the X axis


However, the book solution (what the teacher wants) is:
1,180 N/C <5.5 degrees above the X axis
 
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Your answer is right. However, your solution is unnecessarily complicated. It should only take a few steps:

E1_mag=kq/r1^2
E1_x=E1_mag*cos(theta)
E1_y=E1_mag*sin(theta)

E2_x=kq/r2^2

Ex=E1_x+E2_x
Ey=E1_y
 
So you think I should ignore what the book says? Would certainly make my life a heck of a lot easier.

As for the complication, we're doing the same maths only mine is in a lot more steps - I do that because of a touch of dyslexia, when I swap two numbers I shouldn't its easier to fix it later when every step is down on paper.
 

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