Point Charge Electric Field Problem

In summary: So basically, ignore the book, and just do the following:E1_mag=kq/r1^2E1_x=E1_mag*cos(theta)E1_y=E1_mag*sin(theta)In summary, the student is trying to solve a point charge problem, but is having trouble because the book solution is unnecessarily complicated. Their own solution is to use the Pythagorean theorem, the electric field at (5cm,10cm), 5cm right of the upper charge, and the point charge field equation.
  • #1
Rettro
4
0
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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  • #2
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
 
  • #3
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.
 

Related to Point Charge Electric Field Problem

1. What is a point charge electric field problem?

A point charge electric field problem is a type of physics problem that involves calculating the electric field created by a single point charge at a given distance. This field is a measure of the force that another charge would experience if placed in the vicinity of the point charge.

2. How do I calculate the electric field of a point charge?

To calculate the electric field of a point charge, you can use the formula E = kQ/r^2, where E is the electric field, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), Q is the charge of the point charge, and r is the distance between the point charge and the location where the electric field is being measured.

3. What are the units for electric field in a point charge problem?

The units for electric field in a point charge problem are newtons per coulomb (N/C). This unit represents the amount of force per unit charge that another charge would experience at a given point in space.

4. How does the electric field change as the distance from a point charge increases?

The electric field decreases as the distance from a point charge increases. This is because the force between two charges is inversely proportional to the square of the distance between them, meaning that the electric field at a farther distance is weaker than at a closer distance.

5. Can the electric field of a point charge be negative?

Yes, the electric field of a point charge can be negative. This indicates the direction of the force that another charge would experience if placed at that point. A negative electric field means that the force would push the charge away from the point charge, while a positive electric field would pull the charge towards the point charge.

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