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Tema [17]
3 years ago
9

A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \

rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.

Physics
1 answer:
Rashid [163]3 years ago
4 0

Answer:

the answers are provided in the attachments below

Explanation:

Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface

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Answer:

\bold {1s^22s^22p^63s^23p^64s^23d^2}

Explanation:

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The electronic configuration of Titanium is

1s^22s^22p^63s^23p^64s^23d^2

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A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid
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Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=\frac{6}{100}=0.06 m

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We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

dF=IBdr

dF=IBdr

\tau=rdF=IBrdr

\tau=\int_{0}^{R}IBr dr

\tau=IB(\frac{R^2}{2}

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Where friction force=F

R\times F=\frac{IBR^2}{2}

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F=\frac{3\times 1\times 0.06}{2}

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A positively charged light metal ball is suspended between two oppositely charged metal plates on
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Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

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