Problem 87 - University Physics with Modern Physics

Question

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L .\) The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\) \(\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L .\) (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x .\) (c) Find the electric potential \(V(x)\) in the rod as a function of \(x .\) (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\) .

Step-by-Step Solution

Answer

(a) Total resistance: \( \frac{\rho_0 L}{A}(1-e^{-1}) \), Current: \( \frac{V_0 A}{\rho_0 L (1-e^{-1})} \). (b) \( E(x) = \frac{V_0}{L} e^{-x/L} \). (c) \( V(x) = V_0 (1 - e^{-x/L}) \).

Step 1: Express Differential Resistance

The resistance of an infinitesimal element of the rod, located at position \( x \), can be expressed using the formula \( dR = \frac{\rho(x)\,dx}{A} \). Given \( \rho(x) = \rho_0 \exp(-x/L) \), substitute into the resistance formula: \[ dR = \frac{\rho_0 \exp(-x/L)\,dx}{A} \].

Step 2: Integrate to Find Total Resistance

To find the total resistance \( R \) of the rod, integrate the expression from Step 1 along the length of the rod from \( x=0 \) to \( x=L \):\[ R = \int_0^L \frac{\rho_0 \exp(-x/L)\,dx}{A} \].This simplifies to \[ R = \frac{\rho_0}{A} \left( -L \right)\left[ \exp(-x/L) \right]_0^L \],\[ R = \frac{\rho_0 L}{A}(1 - e^{-1}) \].

Step 3: Calculate Current Using Ohm's Law

According to Ohm's law, the current \( I \) flowing through the rod is given by \( I = \frac{V}{R} \), where \( V = V_0 \) is the potential difference. Substitute \( R \) from the previous step:\[ I = \frac{V_0}{(\rho_0 L / A)(1 - e^{-1})} = \frac{V_0 A}{\rho_0 L (1 - e^{-1})} \].

Step 4: Define Electric Field

The electric field \( E(x) \) is related to the potential drop and resistivity by \( E(x) = \frac{V_0}{L} + \frac{V_0}{L} e^{-x/L} \), considering the gradient of the potential across the rod. Recall \( E(x) = \frac{\rho(x)I}{A} \).

Step 5: Derive Expression for Potential

The electric potential \( V(x) \) is obtained by integrating the electric field. Starting from \( V(x) = -\int_0^x E(x')\,dx' + V_0 \), use \( E(x') = \frac{V_0}{L} e^{-x'/L} \). Integrate to find:\[ V(x) = V_0 (1 - \exp(-x/L)) \].

Step 6: Graph Functions

For graphing purposes, notice that:- \( \rho(x) = \rho_0 \exp(-x/L) \) decreases exponentially from \( \rho_0 \) at \( x=0 \) to \( \rho_0 e^{-1} \) at \( x=L \).- \( E(x) \) being proportional to the current, initially has a stronger negative gradient at the start, reducing as \( x \) increases.- \( V(x) \) starts at \( V_0 \) and approaches zero exponentially as \( x \) increases.Graph these functions with respect to \( x \) to visualize their behavior.

Key Concepts

Ohm's Law

Ohm's Law is one of the fundamental principles in understanding electrical circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. Mathematically, it is expressed as:

• \( I = \frac{V}{R} \)
• Where \(I\) is the current, \(V\) is the voltage, and \(R\) is the resistance.

This relationship tells us that if we have a fixed resistance, an increase in the voltage will result in an increase in current, and vice versa.
For a semiconductor rod with resistance, as in our exercise, knowing the resistivity \(\rho(x)\) helps us calculate the overall resistance, which then lets us determine the current using Ohm's Law.
In cases where resistivity varies along the length of the material, it is critical to account for this variation to accurately calculate the total resistance and understand how the current will flow in the circuitry.

Electric Field

The electric field \(E(x)\) within a material is a crucial concept that helps us understand how the potential energy related to electric charges is distributed within an object.

• The electric field describes the force that a positive test charge would experience within the field.
• In the semiconducting rod mentioned, \(E(x)\) is obtained from the distribution of potential along the rod.

In our exercise, the electric field \(E(x)\) is calculated using the gradient of the potential difference, expressed as a function of \(x\), the position along the rod:

• \( E(x) = \frac{V_0}{L} + \frac{V_0}{L} e^{-x/L} \).

This expression indicates how the electric field starts strong near the potential source and diminishes along the length of the rod due to resistivity.

Electric Potential

Electric potential \(V(x)\) is essentially the potential energy per unit charge at a position \(x\) within an electric field.

• In simple terms, it describes how much potential energy a charge would have at a specific location in an electric field.
• It’s calculated by integrating the electric field over a distance.

For the semiconducting rod, the potential \(V(x)\) as a function of \(x\) can be described as:

• \( V(x) = V_0 (1 - \exp(-x/L)) \).

This equation shows an exponential decay of potential from an initial value \(V_0\) at \(x=0\), approaching zero as \(x\) reaches the other end of the rod at \(x=L\).
Understanding electric potential is pivotal for determining how energy is transformed and transferred in such electronic materials.

Current Calculation

Current calculation in a circuit involves finding out how much charge flows over a point in a circuit per unit time. Using our example, the current \( I \) flowing through the semiconductor rod can be calculated once the total resistance and potential difference are known.

• The step-by-step solution walked through integration to find the resistance of the entire rod.
• With that resistance, we then applied Ohm's Law.

Given by the formula:

• \( I = \frac{V_0 A}{\rho_0 L (1 - e^{-1})} \).

This formula highlights the intricate balance of material properties (area \(A\), resistivity \(\rho_0\), length \(L\)) and the external voltage \(V_0\) dictating the flow of current.
Accurately calculating current is essential for assessing reliability and function in electronics, particularly where material properties vary across a component.