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dalvyx [7]
3 years ago
11

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

R away from the wire using Gauss's Law for Electrostatics.
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

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Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.
mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, \rho=10^{-6}\ \Omega -m

Density, d=8.31\times 10^3\ kg/m^3

(A) The length of the wire is given by using the definition of resistance as :

Volume,

V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3

Area,

A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}....(1)

R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m

(b)  Equation (1) becomes :

A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m

Hence, this is the required solution.                                                                  

5 0
4 years ago
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Romashka [77]
This is not a HES tip
6 0
3 years ago
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A high frequency sound will have a ?
stepladder [879]

Answer:

The frequency of a sound wave is what your ear understands as pitch. A higher frequency sound has a higher pitch and the lower the period

7 0
3 years ago
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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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4 0
2 years ago
Match the indices of refraction with the corresponding effects on the waves.
Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the  speed of light in a vacuum and the speed of light in that medium:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

v=\frac{c}{n}

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 ,n_2 are the index of refraction of the 1st and 2nd medium

\theta_1,\theta_2are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

So we see that if the index of refraction of the second medium is lower (n_2), then the ratio \frac{n_1}{n_2} is larger than 1, so the angle of refraction is larger than the angle of incidence:

\theta_2>\theta_1

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

3 0
4 years ago
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