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jok3333 [9.3K]
3 years ago
9

Consider a long cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at distan

ce r from the axis where r < R. (Use ε0, rho, R, and r as necessary.)
Physics
1 answer:
ahrayia [7]3 years ago
3 0

Answer: Ok, first lest see out problem.

It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.

Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0

where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.

So Q= rho*volume= pi*r*r*L*rho

so replacing : E = (1/2)*r*rho/ε0

you may ask, ¿why dont use R on the solution?

since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.

R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.

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Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee
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<h2>8.0995×10^-21 kgms^-1</h2>

Explanation:

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m_P=1.67\times 10^-^2^7\:kg\\

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Read 2 more answers
A piston/cylinder contains 2 kg of water at 20◦C with a volume of 0.1 m3. By mistake someone locks the piston, preventing it fro
morpeh [17]

Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T=\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

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Explanation:

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