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Charge of uniform surface density (4.0 nC/m2 ) is distributed on a spherical surface (radius = 2.0 cm). What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?

Respuesta :

cjmejiab

To solve this problem we will use the values of the density on surface, which relates the total load and the area of the surface. From there we will get the total charge, which will allow us to find the electric flow.

Surface density is defined as,

[tex]\mu= \frac{Q_{tot}}{A_{surface}}[/tex]

Where,

[tex]A_{surface} = 4\pi r^2[/tex]

Replacing,

[tex]4*10^{-9} = \frac{Q_{tot}}{(4\pi 0.02^2)}[/tex]

[tex]Q_{tot}= 20.11*10^{-2} C[/tex]

At the same time electric flux can be defined as,

[tex]\Phi = \frac{Qtot}{\epsilon_0}[/tex]

Here,

[tex]\epsilon_0 =[/tex] Permittivity vacuum constant

Replacing,

[tex]\Phi =\frac{(20.11 * 10-12 )}{(8.85 * 10-12)}[/tex]

[tex]\Phi =2.27N \cdot m^2 \cdot C^{-1}[/tex]

Therefore the total electric flux through the concentric spherical surface is [tex]2.27Nm^2/C[/tex]

Cricetus

Through conc. spherical surface, the total electric flux will be:

"2.27 N.m².C⁻¹".

Electric flux

Uniform surface density charge, [tex]A_{surface}[/tex] = 4.0 nC/m²

Radius, r = 2.0 cm, or

               = 0.02

We know the relation,

→ Surface density, μ = [tex]\frac{Q_{tot}}{A_{surface}}[/tex]

Here, [tex]A_{surface}[/tex] = 4πr²

                [tex]Q_{tot}[/tex] = 20.11 × 10⁻² C

Now,

→ Electric flux, [tex]\Phi[/tex] = [tex]\frac{Q_{tot}}{\epsilon_0}[/tex]

By substituting the values,

                           = [tex]\frac{20.11\times 10^{-12}}{8.85\times 10^{-12}}[/tex]

                           = 2.27 N.m².C⁻¹

Thus the above response is correct.

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