I should have put -0.09^2 for the distance for the +2.0uC equation. I intuitively knew it was going to be a the sum of the fore magnitudes and in the negative direction. Thank you very much for your help! Greatly appreciated!
Ok, so when I do that I can visualize the center, but not know its physical location. So now that I see center it looks as though from the left bottom vertex I can calculate the height from base to midpoint using tan30 (0.5 x 0.156m) = 0.045 m? And Q2 and Q3 are 0.09 m from vertex to midpoint...
Okay, I do not know ho to find the middle point of the triangle. But once I figure that out am I correct in thinking the net E will be
E = k(+2uC)/distance top vertex to center^2 + E = (2)(sin30)(k-4uC/distance from bottom apex to center^2 = ? I see that both E fields move -y so their...
y/SUB]Homework Statement
Determine the electric field at the center of the triangle. All sides are 15.6 cm. Vetex is 2.0 uC (Q1) while the bottom points are both -4.0 uC (Q2 and Q3).
Homework Equations
E = kQ/r^2
E = F/q
The Attempt at a Solution
Ecenter = E1 + (2)E2 = ?
First calculate...
Sorry, the other answer was not a negative it was 0.01757. I also realized that I was looking at the problem the wrong way and that the distance from Q1 to Q3 was x while the distance from Q2 to Q3 was x - 0.03.I understand now. Thank you for your help!
You calculations reveal x = 0.1024 m which when using your formula provides F = 190.8 N +/- x direction. Now if only I could understand why... Thanks for your help! It is very appreciated.
I just can't understand how this reflects the =0 equation. Isn't the 2kqQ over (0.03 + x)^2 because the 2q acts on Q from the distance 0.03 + x? If you somehow rearranged 2kqQ/(0.03 + x)^2 - kqQ/x^2 = 0 to 2kqQ/x^2 - kqQ/(0.03 - x)^2 = 0 can you please explain how you did it and why you...
Thank you. So I put k(2q)(Q)/(0.03 + x)^2 = - k(-q)(Q)/x2. This allowed elimination of Q and k. I then worked it to (2q)(X^2)/-q = (0.03 + x)^2 to eliminate q resulting in -2x^2 = -(0.03 + x)^2. After factoring the negative through the brackets and using FOIL I come up with the quadratic 3x^2...
Intuitively I can expect the charge would have to be placed to the right of the -q charge. Far enough to the right that the force (due to inverse square of the distance from the 2q charge) will be equal and opposite to the force of the -q charge, with the +2q charge exacting a force in the +x...
Homework Statement
[/B]
A charge of +2q (a) is placed at the origin and a second charge of -q (b) is placed at x = 3.0 cm. Where can a third charge +Q (c) be placed so that it experiences a zero net force?Homework Equations
F = k(Q1)(Q2)/r^2
E = kQ/r^2
Quadratic?The Attempt at a Solution
[/B]...
Thank you very much! I forgot about this being a "fixed position" question. My thoughts on the -/+ x direction components for "q" were based on the forces between "q" and the bottom particles. Would they not also produce a force at an angle towards "q" as they do towards the middle particle? I...
Homework Statement
In this diagram, the net force on the 1.0 mC charge is zero. What is the sign and magnitude of the unknown charge q? Please see attached photo.
[/B]Homework Equations
F = KQ1Q2/r^2[/B]The Attempt at a Solution
I have approached this by first determining the force...