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weqwewe [10]
3 years ago
10

A hollow, conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has a uniform surface charge density

of +6.47 × 10⁻⁶ C/m². A charge of -0.900 μC is now introduced into the cavity inside the sphere.
A. What is the new charge density σ on the outside of the sphere? Express your answer with the appropriate units.
B. Calculate the strength of the electric field E just outside the sphere. Express your answer with the appropriate units.
C. What is the electric flux through a spherical surface just inside the inner surface of the sphere? Express your answer with the appropriate units.
Physics
1 answer:
GarryVolchara [31]3 years ago
5 0
A. What is the new charge density
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The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}

m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

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