A) E along AO
B) 2 E along AO
C) E along BO
D) E along CO
E) zero
Correct Answer: A
Solution :
The situation is as shown below. As shown we resolve E along CO and also along BO into two perpendicular components. The horizontal components cancel each other. The vertical (cosine) components add up along OA to give\[2E\text{ }cos\text{ }60{}^\circ \]. Resultant field along AO \[=2E-2E\cos 60{}^\circ \] \[=2E-2E\times \frac{1}{2}\] \[=2E-E=E\] Hence, resultant field is E along AO.You need to login to perform this action.
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