- #1
chemkid
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Here is my solution. I completed this problem to the best of my ability and I am asking for someone to check and correct my solution if possible. I am not sure if my answers are correct, although and my thinking might also be flawed 
q1= 6.0e-6C
q2= 1.5e-6C
q3=-2.0e-6C
The charges are arranged as such:
O......O...O
q1----0.03m---q2---.02m--q3
The problem is to find the electric field at a point 0.01 meter to the left of q2.
Here is my solution: ( I am ommiting units in calculation for clarity, units are given in answers)
E1y = 0
E1x = Ke q1/r^2 = 8.99e9 * 6.0e-6/.02^2 = 1.35e8 N/C
E2y=0
E2x= Ke q2/r^2 = 8.99e9 * 1.5e-6/.01^2 = 1.35e8 N/C
E3y = 0
E3x = Ke q3/r^2 = 8.99e9 * 2e-6/.03^2 = 2.00e7 N/C
Summing the x components I see that E2x and E1x cancel each other out with respect to the arbitrary point 0.01m to the left of q2. So, I am left with a total electric field of E3 at a point 0.01 cm to the left of q2= 2.00e7 N/C.
Is this correct?
There is a second part of this problem that asks about the force on a point charge placed at the point given its charge of -2.00e-6C. Oddly enough I find finding forces a bit more straightforward. Conceptually I would set it up as a standard charge triangle problem. Seeing as there are no Y components I can just take the sum of the X components in the same way as I did the electric force? Not looking for calculations on the second part of this problem just a conceptual hint on how to go about it, but I would appreciate if someone checks my physics and calculations for the first part of the problem where I have given my answers.
Thanks in Advance,
chemkid
q1= 6.0e-6C
q2= 1.5e-6C
q3=-2.0e-6C
The charges are arranged as such:
O......O...O
q1----0.03m---q2---.02m--q3
The problem is to find the electric field at a point 0.01 meter to the left of q2.
Here is my solution: ( I am ommiting units in calculation for clarity, units are given in answers)
E1y = 0
E1x = Ke q1/r^2 = 8.99e9 * 6.0e-6/.02^2 = 1.35e8 N/C
E2y=0
E2x= Ke q2/r^2 = 8.99e9 * 1.5e-6/.01^2 = 1.35e8 N/C
E3y = 0
E3x = Ke q3/r^2 = 8.99e9 * 2e-6/.03^2 = 2.00e7 N/C
Summing the x components I see that E2x and E1x cancel each other out with respect to the arbitrary point 0.01m to the left of q2. So, I am left with a total electric field of E3 at a point 0.01 cm to the left of q2= 2.00e7 N/C.
Is this correct?
There is a second part of this problem that asks about the force on a point charge placed at the point given its charge of -2.00e-6C. Oddly enough I find finding forces a bit more straightforward. Conceptually I would set it up as a standard charge triangle problem. Seeing as there are no Y components I can just take the sum of the X components in the same way as I did the electric force? Not looking for calculations on the second part of this problem just a conceptual hint on how to go about it, but I would appreciate if someone checks my physics and calculations for the first part of the problem where I have given my answers.
Thanks in Advance,
chemkid