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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Answer:
σ₁ = C/m²
σ₂ = C/m²
Explanation:
Provided Information:
i) Smaller sphere's radius ( r ) = 5 cm.
ii) Larger sphere's radius ( R ) = 12 cm.
iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m
Charge (Q₁) = 572.8 C
Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.
V = constant
∴
=
C
Surface charge density ( σ₁ ) for the larger sphere.
Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ = =
=
C/m².
Surface charge density ( σ₂ ) for the smaller sphere.
Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².
σ₂ = =
=
C/m²