Question

The electric field in a region is given by →E=E0 xl→i. Find the charge contained inside the cubical volume bound by the surfaces x=0, x=a, y=0, y=a, z=0 and z=a. Take E0=5×103 N C-1, l=2 cm and a = 1 cm.

Answer 1

Given :

Length of the cubical volume(l) = 2 cm

E₀ = 5×10³ N/C

a = 1 cm

To Find :

The charge contained in the cubical volume = ?

Solution :

• The electric flux generated in the cubical volume is only due to the surfaces x=0 and x=a.

           $\phi (x=0) =\frac{E_{0}(0)}{l} a^{2}=0$

           $\phi (x=a) =\frac{E_{0}(a)}{l} a^{2}= \frac{E_{0}}{l} a^{3}$

• By using Gauss’s law

            $\frac{q}{\epsilon_{0}} = \phi (x=a)=\frac{E_{0}}{l} a^{3}$

            $q =\frac{E_{0}}{l} a^{3} \epsilon_{0}$

            $q=\frac{5\times10^{-3}}{2\times10^{-2}} \times10^{-6}\times8.854\times10^{-12}$

            $q=2.2\times 10^{-12}$

The charge contained in the cubical volume is 2.2×10⁻¹² C.