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anonymous

  • one year ago

A point charge q1=−4.00nC is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nC is at the point x=0.600 meters, y=0.Calculate the magnitude E of the net electric field at the origin due to these two point charges. Express your answer in newtons per coulomb to three significant figures.

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  1. Astrophysics
    • one year ago
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    |dw:1435733672454:dw|\[E = \frac{ k|q| }{ r^2 }\] you will have to use the following formula.

  2. Astrophysics
    • one year ago
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    |dw:1435734175254:dw| you can make an axis if it makes it easier for you, also to understand if you need a direction as well.

  3. Michele_Laino
    • one year ago
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    here you have to compute a vector sum, in other words the resultant electric field, at the origin of the coordinate system is given by the subsequent formula: \[\Large {\mathbf{E}} = {{\mathbf{E}}_{\mathbf{1}}} + {{\mathbf{E}}_{\mathbf{2}}} = - \frac{{k\left| {{q_1}} \right|}}{{x_0^2 + y_0^2}}{{\mathbf{r}}_{\mathbf{1}}} - \frac{{k{q_2}}}{{x_0^2}}{{\mathbf{r}}_{\mathbf{2}}}\] where r_1 and r_2 are the subsequent vectors: |dw:1435745839643:dw|

  4. Michele_Laino
    • one year ago
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    now, developing that vector equation we get, the subsequent scalar equations: \[\Large \begin{gathered} {E_x} = \frac{{k\left| {{q_1}} \right|}}{{x_0^2 + y_0^2}}\frac{{{x_0}}}{{\sqrt {x_0^2 + y_0^2} }} - \frac{{k{q_2}}}{{x_0^2}} \hfill \\ \hfill \\ {E_y} = - \frac{{k\left| {{q_1}} \right|}}{{x_0^2 + y_0^2}}\frac{{{y_0}}}{{\sqrt {x_0^2 + y_0^2} }} \hfill \\ \end{gathered} \] where E_x and E_y are the components x, y respectively of the resultant vector field, and x_0=0.6 meters, and y_=0.8 meters

  5. Astrophysics
    • one year ago
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    Haha, yup I guess I should've mentioned the vector sum :P, yay @Michele_Laino

  6. Michele_Laino
    • one year ago
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    :) @Astrophysics

  7. anonymous
    • one year ago
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    tnx guys!!

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