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3 years ago

What is the electric flux ΦΦPhi through each of the six faces of the cube?

Physics

1 answer:

denpristay [2]3 years ago

7 0

Flux through each of the six faces of the cube: $\frac{q}{6\epsilon_0}$

Explanation:

In this problem, a charge $q$ is placed at the center of the cube.

We can apply Gauss Law, which states that the flux through the surface of the cube is equal to the charge contained within the cube divided by the vacuum permittivity; mathematically:

$\Phi_T=\frac{q}{\epsilon_0}$

where

$q$ is the charge

$\epsilon_0$ is the vacuum permittivity

Here we want to find the flux through each of the six faces of the cube.

By simmetry, we can say that the 6 faces are identical: therefore, the flux through each of them must be the same. This means that the flux through each faces is 1/6 of the total flux through the total surface, therefore:

$\Phi = \frac{1}{6}\Phi = \frac{q}{6\epsilon_0}$

Learn more about electric fields:

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3 years ago

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Answer:

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3 years ago

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3 years ago

Read 2 more answers

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

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3 years ago

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