Field Intensity at Point X with A(q+) and B(q-): Solve the Question

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At point X, equidistant from a positive charge A (q+) and a negative charge B (q-), the electric field intensity from A is E. When considering the addition of B, the fields must be treated as vectors due to their opposing nature. The correct approach involves vector addition, which means the field from B will partially cancel the field from A. Therefore, the new field intensity at point X is not simply 2E, but rather a resultant value that depends on the magnitudes and directions of the fields. The discussion emphasizes the importance of understanding electric fields as vector quantities rather than scalar sums.
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A point X is located equidistant from point A (q+) and B (q-) in a parallel plane.

If the field intensity of A(q+) at point X is E, then what would the new field intensity be if point B (Q-) is added?

A) 2E
B) 0
C) 1/2E
D) 3/4 E

I assumed the answer is 2E, since the field of the positive charge (A) plus the field of the negative charge (B) are combined, however, my friend says it's B and I'm not sure anymore. Please, help?
 
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Suzukigold said:
A point X is located equidistant from point A (q+) and B (q-) in a parallel plane.

If the field intensity of A(q+) at point X is E, then what would the new field intensity be if point B (Q-) is added?

I assumed the answer is 2E, since the field of the positive charge (A) plus the field of the negative charge (B) are combined …

Hi Suzukigold! :smile:

What do you mean by a parallel plane? :confused:

Any three points are in a plane.

Anyway, electric force is a vector field, so it obeys the law of vector addition, so you can't just add as if they were numbers. :smile:
 
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