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anonymous
 3 years ago
What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?
anonymous
 3 years ago
What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here are the individual electric fields in vector notation: E(1) = kq/r^2 = (8.99e9 N*m^2/C^2)(10e9 C)/(3.0e2 m)^2 = 1.0e5 N/C E(1) = <0, 1.0e5> E(2) = (8.99e9 N*m^2/C^2)(5.0e9 C)/(5.0e2 m)^2 = 1.8e4 N/C E(2) = <1.8e4, 0> r = sqr(3^2 + 5^2) = 5.83 cm E(3) = (8.99e9 N*m^2/C^2)(10e9 C)/(5.83e2 m)^2 = 2.6e4 N/C θ = arctan(3/5) = 30.96° (measured counterclockwise from +x axis) E(3) = <2.6e4cos 30.96°, 2.6e4sin 30.96°> = <2.3e4, 1.4e4>

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1346366361557:dw In the triangle use Pythagoras theorem to find the distance of charge q2 from the point. Then use the formula for Electric field to find E1 and E2 (their respective directions are marked by the arrow heads), then add the fields using parallelogram law of vector addition for the resultant field. Angle between the vector fields has been solved for,\[\theta=tan^{1}2\] PS: Remember to convert the given distances into meters before using them for the field calculation.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0CORRECTION: For the distance between the point and q1, you'd need to use Pythagoras theorem. The distance between the point and q2 is given to be 5cm.
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