Question

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 surface of this cube?

Answer 1

Answer:

The flux through the surface of the cube is $2.314\ Nm^{2}/C$

Solution:

As per the question:

Edge of the cube, 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 calculate the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$                                                      (1)

where

$\phi$ = electric flux

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

Volume Charge density for the given case is given by the formula:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$                  (2)

Volume of cube, $V = a^{3}$

Thus

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

Thus from eqn (2), the total charge is given by:

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

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

Now, substitute the value of 'q' in eqn (1):

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