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Two points, A and B, are separated by 0.0215 m. The potential at A is +89.8 V, and that at B is +38.4 V. Find the magnitude of the constant electric field between the points.

Answer :

Final answer:

The magnitude of the constant electric field between the two points A and B, given the potential at A is +89.8 V, at B is +38.4 V, and the distance between them is 0.0215 m, is approximately 2388 V/m.

Explanation:

The question is asking for the magnitude of the constant electric field between two points, A and B. According to physics, the magnitude of the electric field (E) can be calculated using the formula E = ΔV/ d, where ΔV is the potential difference between the two points and d is the distance between them.

The potential difference (ΔV) can be calculated by subtracting the potential at point B from that at point A, i.e., ΔV = VA - VB = 89.8 V - 38.4 V = 51.4 V. The distance (d) between A and B is given as 0.0215 m.

Substituting these values into the formula gives: E = 51.4 V/ 0.0215 m = approximately 2388 V/m. Therefore, the magnitude of the constant electric field between the two points A and B is roughly 2388 V/m.

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