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anonymous
 one year ago
A point charge of +5.0 micro coulombs is located on the xaxis at x=3.0 cm, and a second point charge of 8.0 micro coulombs is located on the xaxis at x=+4.0 cm. Where should a third charge of +6.0 Micro coulombs be placed so the the the electric field at the origin is zero?
anonymous
 one year ago
A point charge of +5.0 micro coulombs is located on the xaxis at x=3.0 cm, and a second point charge of 8.0 micro coulombs is located on the xaxis at x=+4.0 cm. Where should a third charge of +6.0 Micro coulombs be placed so the the the electric field at the origin is zero?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is the situation described by your problem: dw:1441393802171:dw so starting from this vector equation: \[\Large {{\mathbf{E}}_1}\left( 0 \right) + {{\mathbf{E}}_2}\left( 0 \right) + {{\mathbf{E}}_3}\left( 0 \right) = {\mathbf{0}}\] where all fields are computed at x=0, we get this condition: \[\Large \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}}}{{x_1^2}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_2}}}{{x_2^2}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_3}}}{{{d^2}}} = 0\] where \(\large q_1, q_2, q_3 \) are your charges with their corresponding signs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would \[\frac{ 1 }{ 4\pi \epsilon} \] equal the k constant?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0@ipalnemoatl cool! indeed: \[ k_\text{e} = \frac{1}{4\pi\varepsilon_0}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I figured that was what i had to do, but the problem was that the negatives would cancel everything else. then i realized how the direction of the electric were supposed to go. Thanks.
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