 # A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and beljuA 2021-03-05 Answered

A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and the entire capacitor is 3.0 m long.
a) What is the capacitance per unit length?
b)The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors. Find the charge (magnitude and sign) on both conductors.

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a) Inner radius ${r}_{1}=1.5×{10}^{-3}$ m
outer radius ${r}_{2}=3.4×{10}^{-3}$ m
The capacitance of a cylindrical capacitance is given by
$C=\frac{2\pi {ϵ}_{0}l}{\mathrm{ln}\left(\frac{{r}_{2}}{{r}_{2}}\right)}$

(b)
The inner conductor is at a higher potential than the outer conductor it implies that the sign of the charge on the inner conductor is positive outer conductor is negative the magnitude of charge will be same.
The magnitude of charge on each capacitor plate is,
$q=\left(\frac{C}{l}\right)l{v}_{ab}$

Charge on the outer conductor
Charge on the inner conductor

We have step-by-step solutions for your answer! Jeffrey Jordon

Part(a): The capacitance per unit length of the capacitor is

$1.56×{10}^{-10}F{m}^{-1}$

Part(b): The charge in the inner shell is $+1.6×{10}^{-10}C$ and the charge of the outer shell is $-1.6×{10}^{-10}C$

Explanation:

Part(a):

The capacitance (C) per unit length of a cylindrical capacitor is given by

$C=\frac{2\pi {ϵ}_{0}}{\mathrm{log}\left(b/a\right)}$

where '${ϵ}_{0}$' is the permittivity of free space, 'b' is the radius of the outer shell and 'a' is the radius of the inner shell.

Given, b=3.4mm=0.0034 and a=1.5mm=0.0015 mm. We know, ${ϵ}_{0}=8.854×{10}^{-12}F{m}^{-1}$. Substituting the values in the above equation,

$C=\frac{2\pi ×8.85×{10}^{-12}F{m}^{-1}}{\mathrm{log}\left(0.0034/0.0015\right)}=1.56×{10}^{-10}F{m}^{-1}$

Part(b):

As the voltage on the inner conductor is higher than that of the outer conductor, positive charge resides on the inner shell and negative charge resides on the outer shell.

The charge 'Q' of a capacitor of length 'L' having capacitance 'C' and potential difference 'V' is given by

Q=CVL

Given, $V=350mV=350×{10}^{-3}V$ and L=3.0m. Substituting these values in the above expression

$Q=1.56×{10}^{-10}F{m}^{-1}×350×{10}^{-3}V×3m=1.6×{10}^{-10}C$

The charge in the inner shell is $+1.6×{10}^{-10}C$ and the charge of the outer shell is $-1.6×{10}^{-10}C.$

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