A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

Question

1 month ago

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e of this cube?

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-04T15:26:01+03:00

Answer:

The flux across the cube's surface is $2.314\ Nm^{2}/C$ .

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = $8.0\times 10^{- 2}\ m$ .

Volume charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$ .

Now,

To find the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of vacuum.

The volume charge density for this scenario is described by:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$

Cube volume, $V = a^{3}$ .

Thus,

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$ .

The total charge can be derived from equation (2):

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$ .

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$ .

Now, insert the value of 'q' into equation (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$ .