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anonymous
 4 years ago
how can i calculate the charge on a point q, given the net electric force and the values of 2 other charges?
anonymous
 4 years ago
how can i calculate the charge on a point q, given the net electric force and the values of 2 other charges?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you've got to show the whole question mam

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0You should add the given info into a diagram and/or redraw the diagram.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe this will help ;)

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0Yup. I can't really much though. I can't remember much from this,=.

NotTim
 4 years ago
Best ResponseYou've already chosen the best response.0Eashmore, can you do anything here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We have to balance forces. We know the expression for electromagnetic force is\[F_E = k_e {q_1 q_2 \over r^2}\] First, let's find the component of the net force that acts in the x and ydirections. \[F_{x,3} = F_3 \cos(\theta)\]\[F_{y,3} = F_3 \sin(\theta)\] We note that both forces act positive in their respective directions. Since \(q_1\) is positive and located to the left of \(q_3\), \(q_3\) must be positive if the force from \(q_1\) on \(q_3\) is to be in the positive xdirection. The same thought process verifies this when relating \(q_2\) and \(q_3\). Additionally, let's note that the force from \(q_1\) acts on \(q_3\) solely in the xdirection, and \(q_2\) acts on \(q_3\) solely in the ydirection. Let's balance the forces in x and ydirections now. \[F_{x,3} = k_e {q_1 q_3 \over r^2}\]and\[F_{y,3} = k_e {q_2 q_3 \over r^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok this makes sense so far

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You should be able to rearrange the last two equations for \(q_3\). You should get the same value.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean? set them equal? or factor q3 out of the equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Pick one and solve it for \(q_3\). If we solved both for \(q_3\), both equations should produce the same value for \(q_3\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Fx^2 + Fy^2 = 30.187 ^2 ? and sub both equations in for Fx and Fy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have two unknowns in the above equations yes? so how do I solve for Q3?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0You only have one unknown: \( q_3 \). You know q1 and q2; you know r for each equation; and you know the constant \( k_e \). So you only need to solve for q3.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0(Notice the r=1 m for one of the forces, and r=2m for another force)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0In other words, for both of the equations .... the only variable you don't know is q3. So you can use either equation to solve for q3.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0talk to me ... what's not making sense?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0ok ...I'm out here of here then.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i do not know Fx or Fy I only know the sum of both..sorry having internet problems

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You know \(F_x\) and \(F_y\). Take the total force as being \(F_3\), then use the trigonometric relations I gave you for \(F_x\) and \(F_y\). \[F_x = F_3 \cos(\theta) ~ {\rm and} ~ F_y = F_3 \sin(\theta)\]
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