Problem 77 - 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: \( \frac{\rho_0 L}{A} (1 - \exp(-1)) \), Current: \( \frac{V_0 A}{\rho_0 L (1 - \exp(-1))} \). (b) \(E(x) = \frac{V_0 \exp\left(-\frac{x}{L}\right)}{L(1 - \exp(-1))}\). (c) \(V(x) = V_0 \left(1 - \frac{1 - \exp\left(-\frac{x}{L}\right)}{1 - \exp(-1)}\right)\). (d) Graphs decline exponentially for \(\rho(x)\) and \(E(x)\); linear decay for \(V(x)\).

Step 1: Understanding Resistivity

The resistivity of the rod varies with position and is given by \( \rho(x) = \rho_0 \exp\left(-\frac{x}{L}\right) \), where \( \rho_0 \) is the resistivity at \( x = 0 \). This indicates that resistivity decreases exponentially along the rod from \( x = 0 \) to \( x = L \).

Step 2: Calculate Resistance

To find the total resistance \( R \) of the rod, consider a small element of length \( dx \) at position \( x \). The differential resistance \( dR \) is given by \( dR = \frac{\rho(x)}{A} dx = \frac{\rho_0 \exp\left(-\frac{x}{L}\right)}{A} dx \). Integrating from 0 to \( L \), we get:\[R = \int_0^L \frac{\rho_0 \exp\left(-\frac{x}{L}\right)}{A} dx = \left[ -\frac{\rho_0 L}{A} \exp\left(-\frac{x}{L}\right) \right]_0^L = \frac{\rho_0 L}{A} (1 - \exp(-1))\].

Step 3: Calculate Current

Since the potential difference across the rod is \( V_0 \) and the resistance is \( R \), we use Ohm's law to find the current \( I \). Ohm's law states \( I = \frac{V_0}{R} \), so substitute \( R \) from Step 2 to get:\[I = \frac{V_0 A}{\rho_0 L (1 - \exp(-1))}\].

Step 4: Determine Electric Field Magnitude

The electric field \( E(x) \) is defined as \( E(x) = \frac{V(x)}{L} \). Leveraging Ohm's law, \( E(x) = \frac{I \rho(x)}{A} \). Substituting the expression for \( I \) and \( \rho(x) \) from earlier, we obtain:\[E(x) = \frac{V_0 \exp\left(-\frac{x}{L}\right)}{L (1 - \exp(-1))}\].

Step 5: Find Electric Potential

The potential difference \( dV(x) \) over a small segment \( dx \) is given by \( dV = -E(x)\,dx \). Integrating to find \( V(x) \), we have:\[V(x) = V_0 - \int_0^x E(x')\,dx' = V_0 - \int_0^x \frac{V_0 \exp\left(-\frac{x'}{L}\right)}{L (1 - \exp(-1))}\,dx'\].This simplifies to:\[V(x) = V_0 \left( 1 - \frac{1 - \exp\left(-\frac{x}{L}\right)}{1 - \exp(-1)} \right)\].

Step 6: Graphing Functions

To graph \( \rho(x) \), \( E(x) \), and \( V(x) \) from \( x = 0 \) to \( x = L \), follow these general shapes:- \( \rho(x) \) decreases exponentially from \( \rho_0 \) to \( \rho_0 \exp(-1) \).- \( E(x) \) follows a similar declining exponential trend consistent with \( E(x) = \frac{V_0 \exp\left(-\frac{x}{L}\right)}{L (1 - \exp(-1))} \).- \( V(x) \) starts at \( V_0 \) and decreases towards zero, reaching 0 at \( x = L \).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics, describing the relationship between voltage, current, and resistance in an electrical circuit. It is formulated as \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. This law means that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. For semiconductors like the rod in our exercise, understanding Ohm's Law allows us to calculate the current flowing through the rod when a voltage \( V_0 \) is applied, using the formula \( I = \frac{V_0}{R} \). This equation helps us understand how changes in voltage or resistance will affect the current, which is crucial for designing and analyzing electrical circuits.

Electric Field

The Electric Field \( E(x) \) describes the force per unit charge that a charge would experience at any point \( x \) in the field. In the context of our semiconductor rod, it's the result of a potential difference across the rod. The field can be determined by the change in electric potential \( V \) divided by the distance \( L \): \( E(x) = \frac{V(x)}{L} \).

In our example, knowing that the rod obeys Ohm's Law, the electric field can also be related to the resistivity \( \rho(x) \) of the material and the current \( I \) flowing through it: \( E(x) = \frac{I \rho(x)}{A} \). This relationship shows that the electric field strength is influenced by both the resistivity and the current, highlighting how variations in material properties and current can impact the field within the material.

Electric Potential

Electric Potential \( V(x) \) at any point along our rod is a measure of the potential energy per unit charge at that point. In practical terms, it represents the ability to do work due to the position within an electric field. For the semiconductor rod, the electric potential decreases along its length, starting from \( V_0 \) at \( x = 0 \) and dropping to a value at \( x = L \). The formula to calculate \( V(x) \) when \( E(x) \) is known is \( V(x) = V_0 - \int_0^x E(x') dx' \).

This integral approach considers how the electric field varies across the rod, allowing us to see how potential energy is converted along the rod’s length. Understanding \( V(x) \) is essential for tasks such as determining how much work would be required to move a charge from one point to another within the field.

Exponential Decay

Exponential Decay is a process by which a quantity decreases at a rate proportional to its current value. In the case of the semiconductor's resistivity \( \rho(x) = \rho_0 \exp\left(-\frac{x}{L}\right) \), it describes how the resistivity of the rod changes from \( x = 0 \) to \( x = L \).

This exponential function indicates that resistivity decreases quickly at first, then more gradually, as \( x \) increases. Such behavior is typical in materials where the presence of impurities or other factors cause the conductivity to increase as we move along the rod. Grasping how exponential decay functions is key for predicting material properties over a given distance or time. It’s a crucial tool for engineers and scientists working with materials that change properties rapidly across dimensions, like semiconductors.

Resistance Calculation

Resistance Calculation is an essential part of analyzing any electrical component. For our semiconductor rod with variable resistivity, the total resistance \( R \) is found by integrating over its length. We use the formula for differential resistance, \( dR = \frac{\rho(x)}{A} dx \), and integrate from \( x = 0 \) to \( x = L \) to find \( R \).

The completed integration yields \( R = \frac{\rho_0 L}{A} (1 - \exp(-1)) \), indicating how the exponential decay in resistivity affects the overall resistance. This calculation is vital for scenarios where material properties vary spatially, ensuring accurate predictions about how currents will behave and enabling proper circuit function.

• Understanding the resistance allows for correct application of Ohm's Law to find current.
• It highlights how varying resistivity impacts overall resistance and performance of the material.

This concept is integral to designing reliable and effective electronic devices.