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BARSIC [14]
3 years ago
7

A cube with sides of area 48 cm^2 contains a 28.7 nanoCoulomb charge. Find the flux of the electric field through the surface o

f the cube in unis of Nm^2/C.
Please conceptually explain this question answer to me! Thanks!!
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
5 0

Answer:

The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

Explanation:

Given that,

Area of cube = 48 cm²

Charge = 28.7 nC

We need to calculate the flux of the electric field through the surface

Using formula Gauss's law

The electric flux through any closed surface,

\phi =\dfrac{q}{\epsilon_{0}}

Where, q = charge

Put the value into the formula

\phi=\dfrac{28.7\times10^{-9}}{8.85\times10^{-12}}

\phi =3.24\times10^{3}\ Nm^/C

Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

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Answer:

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Explanation:

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Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

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v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

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v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

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