Problem 55 - University Physics with Modern Physics

Question

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by $$V(x) = Cx^{{4}/{3}}$$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of \(C\). (b) Obtain a formula for the electric field between the electrodes as a function of \(x\). (c) Determine the force on an electron when the electron is halfway between the electrodes.

Step-by-Step Solution

Answer

(a) Calculate \( C \) using \( 240 = C \times (13.0)^{4/3} \). (b) Electric field: \( E(x) = -\frac{4}{3}Cx^{1/3} \). (c) Calculate \( F \) using \( F = qE(6.5) \).

Step 1: Understand the equation for potential

The potential function is given by \( V(x) = Cx^{4/3} \). We need to relate this equation to the given potential difference between the electrodes, which is 240 V, when they are spaced 13.0 mm apart.

Step 2: Determine the value of C

Since the maximum potential difference occurs at the anode (13.0 mm from the cathode), we have: \( V(13.0) = 240 \). Plugging into the potential formula gives \( 240 = C \, (13.0)^{4/3} \). Solve this for \( C \): \[ C = \frac{240}{(13.0)^{4/3}} \]. Compute \( C \) using a calculator to get its value.

Step 3: Derive the electric field as a function of x

The electric field \( E(x) \) is the negative derivative of the electric potential \( V(x) \) with respect to \( x \). Thus, \( E(x) = -\frac{dV}{dx} = -\frac{d}{dx}[Cx^{4/3}] \). Compute this derivative: \( E(x) = -\left(\frac{4}{3}Cx^{1/3}\right) \). So, the electric field as a function of \( x \) is \( E(x) = -\frac{4}{3}Cx^{1/3} \).

Step 4: Calculate the force on an electron halfway between electrodes

Halfway between the electrodes is at \( x = \frac{13.0}{2} = 6.5 \) mm. The electric field at this point is \( E(6.5) = -\frac{4}{3}C(6.5)^{1/3} \). The force on an electron is given by \( F = qE \), where \( q \) is the charge of the electron (\(-1.6 \times 10^{-19} \) C). Therefore, \( F = (-1.6 \times 10^{-19})(E(6.5)) \). Calculate \( E(6.5) \) using the expression from the previous step and multiply by the charge of the electron to find the force.

Key Concepts

Electric Field Derivation

In a cylindrical diode, it is crucial to understand how the electric field is derived from the electric potential. The relationship between electric field and potential is foundational to electromagnetism.
The electric potential given by the function \(V(x) = Cx^{4/3}\) can be differentiated to find the electric field. The electric field \(E(x)\) is the negative derivative of the potential \(V(x)\) with respect to \(x\).
To derive \(E(x)\):

• Take the derivative \(\frac{d}{dx}[Cx^{4/3}]\), applying the power rule.
• This results in \(E(x) = -\frac{4}{3}Cx^{1/3}\).

The negative sign emphasizes that the electric field direction is from high to low potential. This computation helps in understanding how potential differences translate into electric field strengths between the electrodes.

Force on an Electron

Analyzing the force on an electron between the electrodes allows us to understand how charges move in an electric field. An electron, having a negative charge, will experience a force when placed in this field.
The formula to calculate this force is given by \(F = qE\), where \(q\) is the charge of the electron and \(E\) is the electric field at a specific point.
Here’s how to do it:

• First, we find \(E(x)\) for \(x = 6.5\) mm, which is halfway between the diodes.
• Use the derived formula: \(E(6.5) = -\frac{4}{3}C(6.5)^{1/3}\).
• Then, multiply \(E(6.5)\) by the charge of the electron, \(-1.6 \times 10^{-19}\) C.

This gives the force \(F\) on the electron. The negative sign indicates the direction is opposite to the field.

Vacuum Tube Diode

A vacuum tube diode is an electronic component used to control electrical current. Comprising two electrodes, a cathode, and an anode, these components are crucial in allowing current to flow in a single direction.
- The cathode emits electrons when heated.
- The anode collects these electrons, establishing a flow of current.
Its cylindrical design enhances uniformity in the electric field distribution, which is crucial for effective functionality. The space between electrodes must accommodate a sufficient potential difference, ensuring proper electron flow across the vacuum.

Potential Difference in Diodes

In cylindrical diodes, the potential difference between the cathode and the anode is vital for diode operation. It determines the electron flow efficiency from the cathode to the anode.
The potential difference is the energy required to move a charge between two points in an electric field. For this exercise, it’s defined as 240 V across a given distance of 13.0 mm.
To calculate the constant \(C\) in the potential equation \(V(x) = Cx^{4/3}\), use:

• Set \(V(13.0) = 240\) and solve: \[C = \frac{240}{(13.0)^{4/3}}\].
• This relationship defines how potential varies over distance \(x\).

The value of \(C\) uniquely characterizes the diode under specific operating conditions, playing a critical role in ensuring the diode functions as intended.