Physics

Current Electricity

Current density

Two large parallel metal plates are located in vacuum. One of

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Hard

Solving time: 4 mins

Two large parallel metal plates are located in vacuum. One of these serves as a cathode, a source of electrons with negligible initial speed. An electron flow directed toward the opposite plate produces a space charge causing the potential in the gap between the plates to vary as $V = a x^{4/3}$ , where $a$ is a positive constant, and $x$ is the distance from the cathode. Find: (a) The volume density of space charge as a function of $x$ (b) The current density.

Views: 5,576 students

Updated on: Jun 15, 2023

Book:

Physics Galaxy Electrostatics and Current Electricity Vol 3 A (Ashish Arora)

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Levi Discussed

Two large parallel metal plates are located in vacuum. One of these serves as a cathode, a source of electrons with negligible initial speed. An electron flow directed toward the opposite plate produces a space charge causing the potential in the gap between the plates to vary as

V = a x^{4/3}

, where

a

is a positive constant, and

x

is the distance from the cathode. Find: (a) The volume density of space charge as a function of

x

(b) The current density.

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Solution (a) Given that the potential is given as

V = a x^{4/3}

Using above potential, the electric field in space is given as

E = - \frac{d V}{d x} = - \frac{4 a}{3} x^{1/3}

Using above electric field, volume density of space charge can be obtained by Gauss's law in the region for a small cylindrical Gaussian surface of cross section S and width dx, applied as

d E . S = \frac{ρ ( x ) S d x}{\in _{0}}

\Rightarrow - \frac{4 a}{3} (\frac{1}{3} x^{- 2/3} d x) S = \frac{ρ ( x ) S d x}{ϵ _{0}}

\Rightarrow ρ (x) = - \frac{4}{9} a \in_{0} x^{2/3}

(b) Current density in the region is given as

\Rightarrow J = - \frac{4}{9} a ε_{0} x^{- 2/3} v (x)

Due to the potential gain in kinetic energy of electrons is given as

\frac{1}{2} m v^{2} (x) \Rightarrow v (x) = e V (x) = \frac{2 V ( x ) e}{m}

Substituting above value of

v (x)

in equation-(3.35) gives

\Rightarrow J = - \frac{4}{9} ε_{0} a^{3/2} 2 e / m

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Question Text	Two large parallel metal plates are located in vacuum. One of these serves as a cathode, a source of electrons with negligible initial speed. An electron flow directed toward the opposite plate produces a space charge causing the potential in the gap between the plates to vary as $V = a x^{4/3}$ , where $a$ is a positive constant, and $x$ is the distance from the cathode. Find: (a) The volume density of space charge as a function of $x$ (b) The current density.
Updated On	Jun 15, 2023
Topic	Current Electricity
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 3
Upvotes	394
Avg. Video Duration	5 min

Text solutionVerified

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