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anonymous
 one year ago
How do I solve this problem?
Point charges of 4 μC and 3 μC are placed 0.34 m apart. Where can a third charge be placed so that the net force on it is zero? Give answer in terms of meters from the 3 μC charge.
anonymous
 one year ago
How do I solve this problem? Point charges of 4 μC and 3 μC are placed 0.34 m apart. Where can a third charge be placed so that the net force on it is zero? Give answer in terms of meters from the 3 μC charge.

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johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433975431068:dw

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Now, where can we put a test charge in this diagram so that it will have a net force of 0 acting on it Remember, no matter where we put it...it will feel 2 electrostatic forces Call 4uC q1 3uC q2 and our test charge q3 \(\Large k\frac{q_1 q_3}{(r_{1,3})^2}\) and \(\Large k\frac{q_2 q_3}{(r_{2,3})^2}\) So we need to find out the position that will allow these two to cancel out.

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1So, we put a positive test charge somewhere...but where? Well since positive will repel positive but be attracted to the negative However, since the positive charge is of greater magnitude here...it will repel MORE than the negative charge will ATTRACT So the test charge would need to be placed somewhere over... dw:1433975891188:dw

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1The distance between q2 and q3 can be labeled as 'x' so that means the distance between q1 and q3 would be 0.34 + x dw:1433976136910:dw

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1So...now we have something we can solve \[\large k\frac{q_1 q_3}{(r_{1,3})^2} + k\frac{q_2 q_3}{(r_{2,3})^2} = 0\] Can you continue from here?
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