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Chapter 23: Problem 55

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 \(V(x)=C x^{4 / 3}\) 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.

Short Answer

Expert verified
The constant \(C\) in the given potential equation has a value which is calculated using the given potential difference and distance between the electrodes. The electric field as a function of \(x\) is -\(\frac{4}{3}\)C\(x^{\frac{1}{3}}\). The force exerted on the electron when it is halfway between the electrodes can be obtained by calculating the electric field at that point and substituting it into the formula \(F = e \cdot E\) along with the known charge of an electron.

Step by step solution

01

Calculate the value of C

The equation for electric potential \(V\) is given as \(V(x)=C \cdot x^{4 / 3}\). We also know that at \(x = 13.0 mm\) (or \(0.013 m\)), the potential \(V = 240V\). We can substitute these values into the equation, equate it to \(V\) and solve for \(C\). The solution gives \(C = \frac{V}{{x^{4/3}}}\). Hence, substituting \(V=240 V\) and \(x=0.013 m\) we find the value of \(C\).
02

Derive formula for the electric field

The electric field \(E\) can be obtained by taking the negative derivative of electric potential \(V\) with respect to \(x\). Hence, \(E(x) = -\frac{dV(x)}{dx}\). Substituting \(V(x) = C \cdot x^{4/3}\) into this equation and computing the derivative will give us the equation for electric field in terms of \(x\) and \(C\).
03

Determine the force on the electron

When the electron is halfway between the electrodes, the distance is \(x = 13.0mm / 2 = 0.0065m\). Using the equation obtained from Step 2, we calculate the electric field at this point. The force \(F\) experienced by the electron in the electric field can be computed using the formula \(F = e \cdot E\), where \(e = 1.6 x 10^{-19} C\) is the charge of the electron. So, substituting the computed electric field and the known charge, we find the force on the electron.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential in a vacuum tube diode is a concept about the energy required by a charge to move within an electric field to a certain point. In this exercise, the electric potential between the cathode and anode is given by a function: \(V(x)=C \cdot x^{4 / 3}\). Here, \(C\) is a constant unique to the diode based on its specific operational conditions, and \(x\) is the distance from the cathode. Understanding electric potential provides insight into how energy varies as charged particles move between electrodes. To calculate this potential difference we can use the known distance between the electrodes and the voltage:
  • The given distance between the cathode and anode is \(13.0 \, ext{mm} = 0.013 \, ext{m}\).
  • The potential difference is \(240 \, ext{V}\).
By substituting these values into the formula and solving for \(C\), students find the constant that defines this specific diode's electric characteristics.
Electric Field
The electric field is a vector field that represents the force experienced by a positive test charge placed in the field. It is related to the rate of change of the electric potential with respect to distance. In mathematical terms, within this exercise, the electric field \(E(x)\) is expressed as the negative gradient of the electric potential. The formula is represented as: \(E(x) = -\frac{dV(x)}{dx}\)Substituting the potential function \(V(x) = C \cdot x^{4/3}\) into this derivative equation allows us to derive the electric field's formula. By understanding this relationship, students can see how changes in potential translate to forces on charged particles. This conversion is crucial in predicting how electronic devices like vacuum tubes manage charge movement effectively.
Force on Electron
Force on an electron in an electric field is calculated using the simple formula \(F = e \cdot E\), where \(e\) is the charge of an electron \(1.6 \times 10^{-19} \, ext{C}\). Halfway between the electrodes, the electric field impacts the electron's movement, thereby creating a force. To find the force at this midpoint, consider:
  • Distance when halfway: \(x = 0.0065 \, ext{m}\).
  • Use the previously derived electric field equation to compute \(E\).
By applying the values, students can solve for the force \(F\). This illustrates how the electric field influences electrons, a principle important for understanding the physics driving the functionality of devices like vacuum tube diodes.
Cathode and Anode
Cathode and Anode are terms describing electrodes in a vacuum tube diode, each with a distinct role. The cathode is negatively charged and serves as the source of electrons. Electrons are emitted from the cathode surface due to thermionic emission, a process which releases electrons when the cathode is heated. The anode, in contrast, is positively charged and attracts these electrons towards it. Understanding the relationship between cathode and anode is essential for comprehending vacuum tubes function:
  • The electrons move from the cathode to the anode driven by the electric field.
  • This movement creates a current essential for the diode's operation.
This component naming is intrinsic not just for theory but for practical application in building and understanding electronic circuits.

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Most popular questions from this chapter

A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{~mm}\), and the potential difference between them is \(360 \mathrm{~V}\). (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Two point charges \(q_{1}=\) \(+2.40 \mathrm{nC}\) and \(q_{2}=-6.50 \mathrm{nC}\) are \(0.100 \mathrm{~m}\) apart. Point \(A\) is midway between them; point \(B\) is \(0.080 \mathrm{~m}\) from \(q_{1}\) and \(0.060 \mathrm{~m}\) from \(q_{2}\) (Fig. E.23.19). Take the clectric potential to be zero at infinity. Find (a) the potential at point \(A ;\) (b) the potential at point \(B ;(\mathrm{c})\) the work done by the electric field on a charge of \(2.50 \mathrm{nC}\) that travels from point \(B\) to point \(A\).

Identical charges \(q=+5.00 \mu \mathrm{C}\) are placed at opposite corners of a square that has sides of length \(8.00 \mathrm{~cm}\). Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_{0}=-3.00 \mu \mathrm{C}\) is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(\bar{q}_{0}\) when it is at point \(A ?\) Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_{0}\) when it is at point \(B ?\) (c) How much work does the electric force do on \(q_{0}\) during its motion from \(A\) to \(B ?\) Is this work positive or negative? When it goes from \(A\) to \(B,\) does \(q_{0}\) move to higher potential or to lower potential?

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?
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