A cylindrical capacitor is made of two concentric conducting cylinders. The inner cylinder has radius R1 = 19 cm and carries a uniform charge per unit length of λ = 30 μC m. The outer cylinder has radius R2 = 45 cm and carries an equal but opposite charge distribution as the inner cylinder > 14% Part (a) Use Gauss' Law to write an equation for the electric field at a distance R1 < r < R2 from the center of the cylinders. Write your answer in terms of, r, and eo Grade Summa ry 0% 100% Potential Submissions Attempts remaining:

Gauss Law states the net electric flux that is passing through an enclosed area would have a direct proportionality with the combination of the electric charge enclosed in that area

Mathematically,

∮[tex]E \cdot dA =\frac{q}{\epsilon_0}[/tex]

[tex]E(2 \pi r l)=\frac{q}{\epsilon_0}[/tex]

[tex]E=\frac{q}{2 \pi r l \epsilon_0}[/tex]

[tex]= \frac{\lambda}{2 \pi \epsilon_0 r}[/tex]

The l was eliminated from the equation because R1 < r < R2

Mathematically the potential difference is given as

[tex]V =\int\limits^{R1}_{R2} {E} \, dr[/tex]

[tex]=\int\limits^{R1}_{R2} {\frac{\lambda}{2 \pi \epsilon_0 r}} \, dr[/tex]

[tex]=\frac{\lambda}{2 \pi \epsilon_0 r} (ln(R2)-ln(R1))[/tex]

[tex]\frac{\lambda}{2 \pi \epsilon_0 r}ln(\frac{R2}{R1} )[/tex]

Substituting

[tex]30 \mu C \ m for \ \lambda \ 8.8*10^{-12} \ for \ \epsilon_0 45cm \ for \ R2 \\and \ 19cm \ for \ R1[/tex]

[tex]V =\frac{30*10^{-6}}{2* (3.142)(8.85*10^{-12}))}*ln(\frac{45cm}{19cm} )[/tex]

[tex]= 4.652 *10^5Volt[/tex]

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-02-19T07:26:43+01:00

The electric field is [tex]E=\frac{4.77}{\epsilon_o r}\times10^{-6} \;V/m[/tex]

According to the question there are two concentric cylinders one of radius R₁= 19cm = 0.19m and another of radius R₂= 45cm = 0.45m, having linear charge density of λ = 30μC and λ = -30μC respectively.

Gauss Law:

From Gauss Law, we can calculate the electric field at a distance R₁< r < R₂ as below:

[tex]\int\limits^{}_{ }{E} \, ds=\frac{Q_{enclosed}}{\epsilon_o}[/tex]

Now the enclosed charge inside the Gaussing surface will only be from the inner cylinder since R₁< r < R₂.

The Gaussia surface in this case will be a cylinder of length l and radius r.

Since the cylinder has linear charge density, the total charge enclosed in the Gaussian surface is:

[tex]Q_{enclosed}=\lambda l[/tex]

thus,

[tex]\int\limits^{}_{ }{E} \, ds=\frac{Q_{enclosed}}{\epsilon_o}\\\\E\times2\pi rl=\frac{\lambda l}{\epsilon_o} \\\\E=\frac{\lambda}{2\pi \epsilon_o r} \\\\E=\frac{4.77}{\epsilon_o r}\times10^{-6} \;V/m[/tex]

Learn more about Gauss Law:

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