Rettro
Apr2-11, 05:18 PM
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.
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
Point Charge Field, E = [Kq/r^2] * Unit Vector
Pythagorean Theorem, A^2+B^2=C^2
3. 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
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
Point Charge Field, E = [Kq/r^2] * Unit Vector
Pythagorean Theorem, A^2+B^2=C^2
3. 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