Two cylindrical conductors with equal cross-sections and different resistivites `rho_(1)` and `rho_(2)` are point end to end. Find the charge at the boundary of the conduction if a current `I` flows from conductor 1 to conductor 2

Options

A. `epsilon_(0)(rho_(2)-rho_(1))l`
B. `epsilon_(0)(2rho_(2)+rho_(1))l`
C. `2epsilon_(0)(rho_(1)-rho_(2))l`
D. `epsilon_(0)(rho_(1)+rho_(2))l`

Answer

To solve the problem of finding the charge at the boundary of the two cylindrical conductors when a current \( I \) flows from conductor 1 to conductor 2, we can follow these steps:

### Step 1: Understand the Current Density
Since both conductors have equal cross-sectional areas \( A \) and carry the same current \( I \), the current density \( J \) can be defined as:
\[ J = \frac{I}{A} \]

### Step 2: Relate Electric Field and Current Density
Using Ohm's Law, the electric field \( E \) in each conductor can be expressed in terms of the resistivity \( \rho \) and the current density \( J \):
\[ E_1 = \rho_1 J \quad \text{(for conductor 1)} \]
\[ E_2 = \rho_2 J \quad \text{(for conductor 2)} \]

### Step 3: Apply Gauss's Law
According to Gauss's Law, the integral of the electric field \( E \) over a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon \):
\[ \oint E \cdot dS = \frac{Q_{\text{enclosed}}}{\epsilon} \]
For the boundary between the two conductors, we can write:
\[ E_2 - E_1 = \frac{\sigma}{\epsilon} \]
where \( \sigma \) is the surface charge density at the boundary.

### Step 4: Substitute Electric Fields
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \rho_2 J - \rho_1 J = \frac{\sigma}{\epsilon} \]
This simplifies to:
\[ (\rho_2 - \rho_1) J = \frac{\sigma}{\epsilon} \]

### Step 5: Solve for Surface Charge Density
Rearranging gives us the surface charge density \( \sigma \):
\[ \sigma = \epsilon (\rho_2 - \rho_1) J \]

### Step 6: Calculate the Total Charge at the Boundary
The total charge \( Q \) at the boundary can be found by multiplying the surface charge density \( \sigma \) by the area \( A \):
\[ Q = \sigma A = \epsilon (\rho_2 - \rho_1) J A \]
Substituting \( J = \frac{I}{A} \):
\[ Q = \epsilon (\rho_2 - \rho_1) \left(\frac{I}{A}\right) A \]
This simplifies to:
\[ Q = \epsilon (\rho_2 - \rho_1) I \]

### Final Answer
Thus, the charge at the boundary of the conductors is:
\[ Q = \epsilon (\rho_2 - \rho_1) I \]