Problem 57 - 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 not a linear function of the position, even with planar geometry, but is given by $$V(x)=C x^{4 / 3}$$ 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 \(\mathrm{mm}\) and the potential difference between electrodes is 240 \(\mathrm{V}\) . (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 \(C = \frac{240}{(0.013)^{4/3}}\); (b) \(E(x) = -C \cdot \frac{4}{3} x^{1/3}\); (c) Compute \(F = e \cdot E(0.0065)\).

Step 1: Determine the Constant C

To determine the constant \(C\), we use the potential difference between the cathode and the anode. Given that the distance between the cathode and anode is 13.0 mm (or 0.013 m) and the potential difference is 240 V, we set \( V(x) = C x^{4/3} \). At the anode, where \( x = 0.013 \) m, \( V(x) = 240 \) V.Plugging in these values:\[240 = C (0.013)^{4/3}\]Solving for \(C\):\[C = \frac{240}{(0.013)^{4/3}}\]Calculate the above value to find \(C\).

Step 2: Calculate the Electric Field E(x)

The electric field \(E(x)\) is the negative gradient of the potential \(V(x)\).The formula for the electric field is given by:\[E(x) = -\frac{dV}{dx}\]Calculating the derivative of \(V(x) = C x^{4/3}\):\[\frac{dV}{dx} = C \cdot \frac{4}{3} x^{1/3}\]Thus, the electric field \(E(x)\) becomes:\[E(x) = -C \cdot \frac{4}{3} x^{1/3}\]

Step 3: Determine the Force on the Electron

The force \(F\) on an electron in an electric field \(E(x)\) is given by \(F = e \times E(x)\), where \(e\) is the charge of an electron (\( e = -1.6 \times 10^{-19} \) C).We need to find \(E(x)\) when the electron is halfway between the electrodes. Halfway means \(x = 0.0065\) m.Substitute \(x\) into the equation for \(E(x)\):\[E(0.0065) = -C \cdot \frac{4}{3} (0.0065)^{1/3}\]Finally, calculate the force:\[F = e \cdot E(0.0065) = -1.6 \times 10^{-19} \cdot E(0.0065)\]Evaluate \(E(0.0065)\) and then calculate \(F\).

Step 4: Conclusion

Once all the calculations are performed:1. Determine the value of \(C\) for the given conditions.2. Express the electric field \(E(x)\) as a function of \(x\).3. Compute the force on an electron at \(x = 0.0065\) m using the calculated \(E(x)\).

Key Concepts

Potential Difference

The potential difference in a vacuum tube diode is a measure of how much electrical energy is converted to kinetic energy. It is the voltage between the cathode and the anode. Using the equation provided, where the potential difference is given by \( V(x) = C x^{4/3} \), we find the constant \( C \) by using the known values.
Between the electrodes, the potential difference is 240 V, and this occurs over a distance of 13.0 mm, or 0.013 m. By setting

• \( V(0.013) = 240 \text{ V} \)
• \( C (0.013)^{4/3} = 240 \)

We solve for \( C \) as:
\[ C = \frac{240}{(0.013)^{4/3}} \]
This constant characterizes the specific diode setup, dictating how the potential changes as one moves away from the cathode towards the anode.

Cathode and Anode

In a vacuum tube diode, two main components are crucial: the cathode and the anode.
- **Cathode** is the negatively charged electrode where electrons are emitted. - **Anode** is the positively charged electrode that attracts the electrons emitted by the cathode.
The potential difference between these electrodes sets up a path for electron flow, determining how efficiently the diode operates.
Due to the different charges, electrons naturally move from the cathode to the anode. Initially, this movement faces resistance due to the electric field established between the two electrodes. Therefore, the configuration needs careful consideration to ensure efficient electron transfer.

Electric Field Gradient

The electric field gradient describes how the electric field strength changes with distance between the cathode and anode.
This gradient is essential in determining how much work is done on an electron as it moves. In our scenario, the electric field \( E(x) \) is derived directly from the potential difference equation.
The relation is:

• \( E(x) = -\frac{dV}{dx} \)
• Using \( V(x) = C x^{4/3} \), we find \( \frac{dV}{dx} = C \cdot \frac{4}{3} x^{1/3} \)

Thus, the electric field becomes:
\[ E(x) = -C \cdot \frac{4}{3} x^{1/3} \]
This field points from anode to cathode, indicating the direction of force on a charge, with the negative sign showing that electrons are pushed in the direction opposite the field.

Force on an Electron

The force exerted on an electron in an electric field is a direct result of the field's strength. This force moves electrons from the cathode to the anode, facilitating electrical current.
To compute this force, we use the formula \( F = e \cdot E(x) \), where \( e \) is the electron charge, \( -1.6 \times 10^{-19} \) C, and \( E(x) \) is the electric field.
Specifically, to find the force when an electron is halfway between the electrodes at \( x = 0.0065 \) m, we substitute into the derived \( E(x) \):

• \( E(0.0065) = -C \cdot \frac{4}{3} (0.0065)^{1/3} \)

The resulting force is:
\[ F = -1.6 \times 10^{-19} \cdot E(0.0065) \]
This calculation demonstrates the interaction within the diode, showing how the electric properties translate into mechanical force upon charged particles.