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Bad White [126]
3 years ago
11

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

int r < R1 is:
Physics
1 answer:
gregori [183]3 years ago
6 0

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

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Answer:

3.222 ohms

Explanation:

If the total wire had a resistance of 29 ohms, when cut in three, each piece will have a resistance of 9.666 ohms.

As these three pieces (R1, R2 and R3) are now connected in parallel, the equivalent resistance R can be calculated using this equation:

1/R = 1/R1 + 1/R2 + 1/R3

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1/R = 3/9.666

R = 9.666/3 = 3.222 ohms

The resistance between A and B will be 3.222 ohms

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Consider a mass-spring system. The spring has a spring constant 2.17e 3 N/m. On the end is a mass of 4.71 kg. You start at equil
Novosadov [1.4K]

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