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

[Suzukigold]

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?

 

[tiny-tim]

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!

What do you mean by a parallel plane?

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.

 

[thomate1]

I would approach this question by first understanding the concept of field intensity. Field intensity is a measure of the strength of an electric field at a particular point, and it is directly proportional to the magnitude of the charge creating the field.

In this scenario, we have two charges, A(q+) and B(q-), located at points A and B respectively. Point X is located equidistant from both points A and B. This means that the distance between point X and both charges is the same, and therefore, the field intensity at point X due to both charges will be equal.

Now, to solve the question, we need to consider the direction of the electric fields created by each charge. The field intensity at point X due to A(q+) is E, but the field intensity due to B(q-) will be in the opposite direction. This is because opposite charges create electric fields in opposite directions.

Therefore, the total field intensity at point X will be the sum of the two individual field intensities, which is given by 2E. This is because the field intensity at point X due to B(q-) will have the same magnitude as E, but in the opposite direction. When we add these two intensities together, we get a total field intensity of 2E.

Hence, the correct answer is A) 2E. It is important to consider the direction of the electric fields when calculating the total field intensity at a point. This understanding is crucial in accurately predicting the behavior of electric fields and charges.