Question

Two cylindrical conductors with equal cross-sections and different resistivities ${\rho }_1$ and ${\rho }_2$ are put end to end. Find the charge at the boundary of the conductors if a current $I$ flows from conductor $1$ to conductor $2$ ._​

• A. ${\epsilon }_0\left({\rho }_2-{\rho }_1\right)I$
• B. ${\epsilon }_0\left(2{\rho }_2+{\rho }_1\right)I$
• C. $2{\epsilon }_0\left({\rho }_1-{\rho }_2\right)I$
• D. ${\epsilon }_0\left({\rho }_1+{\rho }_2\right)I$

Answer 1

Answer: (A)

According to ohm's law, $\overrightarrow{E}=\text{ρ}\overrightarrow{J}$
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
$J_1=\frac{I}{\pi R^2}$ (Along the length of the conductor)

Similarly, $J_2=\frac{I}{\pi R^2}$ (along the length of the conductor)
The electric field inside the first conductor is
$E_2=\frac{{\rho }_{2}I}{\pi R^2}$ (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
$S=\pi R^2$ ​
The net flux through this region is
$\text{ϕ}={\text{ϕ}}_1+{\text{ϕ}}_2={\overrightarrow{E}}_1.\overrightarrow{S}+{\overrightarrow{E}}_2\overrightarrow{S}$
$=E_{1}Scos180^{\circ }+E_{2}Scos{0}^{\circ }=-E_{1}S+E_{2}S=(E_2-E_1)S$
According to Gauss's Law,
$\varphi =\frac{q}{{\epsilon }_0}$
or $\left(E_2-E_1\right)=\frac{q}{S{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}$
$\therefore$ The surface charge density at the boundary is
$\sigma =\left(E_2-E_1\right){\epsilon }_0$
Putting the value of $E_1$ and $E_2$, we get
$\therefore \sigma =\left({\rho }_{2}j-{\rho }_{1}j\right)\text{ or }\sigma =\left({\rho }_2-{\rho }_1\right)j$
$\therefore \sigma ={\epsilon }_0\left({\rho }_2-{\rho }_1\right)\frac{I}{\pi R^2}$
or $\sigma \pi R^2={\epsilon }_0\left({\rho }_2-{\rho }_1\right)I$
or $q={\epsilon }_0\left({\rho }_2-{\rho }_1\right)I$
Charge at the boundary $q={\epsilon }_0\left({\rho }_2-{\rho }_1\right)I$