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Tasya [4]
3 years ago
8

Consider a spherical capacitor with radius of the inner conducting sphere a and the outer shell b. The outer shell is grounded (

i.e., it is at zero potential). The charges are +Q and −Q. a A B +Q −Q b What is the magnitude of electric field at in the region between the sphere and the outer

Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

Explanation:

The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.

The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)

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Answer:

15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2

Explanation:

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The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started
bekas [8.4K]

Answer:

-54.12 V

Explanation:

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W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

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A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca
viva [34]

Answer:

a=0.2*10^{-5}g

Explanation:

From the question we are told that:

Mass M=180=>0.18kg

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Length of Wire L=8.6cm=>0.086

Current I=30A

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Therefore

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Therefore in Terms of g's

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