Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed u
niformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
1 answer:
Answer:
The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.
Explanation:
Here’s the breakdown:
The electric field from the positive sheets E₁ = б/2E₀
E₂ is from the negative sheet = -б/2E₀
At points 1, 4, and 5, the electric fields created by the sheets oppose each other.
At point 1, the total field is calculated as -E₁ + E₂ = 0.
Similarly, at point A, the total field results in -E₁ - E₂ = 0.
However, at any point in between the plates, the electric field is directed consistently in one way.
At points 2 and 3, the field is directed to the right.
Thus, we have:
E net = E₁ + E₂
= б/2E₀ + -б/2E₀
=б/E₀
Note: Please refer to the attached document for the full question accompanying this solution.
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