Physics - Current Electricity Question with Solution | TestHub

According to ohm's law,$\overrightarrow{\text{E}}=\text{ρ}\overrightarrow{\text{J}}$

The problem is solved by the following steps:

Step I:Determine the electric field in both conductors.

The current density in the first conductor is

$J_1=\frac{\mathit{\text{I}}}{\text{π}{\mathit{\text{R}}}^2}$(Along the length of the conductor)



Similarly,$J_2=\frac{I}{\text{π}{\mathit{\text{R}}}^2}$(along the length of the conductor)

The electric field inside the first conductor is
${\mathit{\text{E}}}_2=\frac{{\mathit{\text{ρ}}}_2\mathit{\text{I}}}{\text{π}{\mathit{\text{R}}}^{\mathrm{2 }}}$(along the length of the conductor)

Step II:Consider small cylindrical region near the boundary and apply Gauss's law:

We consider a small cylindrical region at the boundary



The area of the cylindrical region is

$S = \mathrm{\pi }{R}^2$​
The net flux through this region is

$\text{ϕ}={\text{ϕ}}_1+{\text{ϕ}}_2={\overrightarrow{\text{E}}}_1.\overrightarrow{\text{S}}+{\overrightarrow{\text{E}}}_2\overrightarrow{\text{S}}$

$= E_{1}S \cos 180° + E_{2}S \cos 0°= - E_{1}S + E_{2}S = (E_2 - E_1) S$

According to Gauss's Law,

$\text{ϕ}=\frac{\text{q}}{{\text{ε}}_0}$

$\text{or}\left({\text{E}}_2-{\text{E}}_1\right)=\frac{\text{q}}{\text{S}{\text{ε}}_0}=\frac{\text{σ}}{{\text{ε}}_0}$

∴ The surface charge density at the boundary is

$\sigma = (E_2 - E_1) {\epsilon }_0$

Putting the value of E₁ and E₂, we get

$\therefore \sigma = ({\rho }_2\mathrm{j} - {\rho }_1\mathrm{j}) or \sigma = ({\rho }_2 - {\rho }_1)\mathrm{j}$

$\therefore \text{σ}={\text{ε}}_0\left({\text{ρ}}_2-{\text{ρ}}_1\right)\frac{\text{I}}{\text{π}{\text{R}}^2}$

or$\sigma \mathrm{\pi }{R}^2= {\epsilon }_0({\rho }_2 - {\rho }_1)I$

or$q = {\epsilon }_0({\rho }_2 - {\rho }_1)I$

Charge at the boundary$q = {\epsilon }_0({\rho }_2 - {\rho }_1)I$