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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A long, straight wire has fixed negative charge with a linear charge density of magnitude 3.6nC/m . The wire is to be enclosed by a coaxial, thin-walled non-conducting cylindrical shell of radius 1.5 cm . The shell is to have positive charge on its outside surface with a surface charge density s that makes the net external electric field zero. Calculate s.

The surface density of the outside surface is $3.8×{10}^{-8}\mathrm{C}/{\mathrm{m}}^{2}$ .

See the step by step solution

## Step 1: The given data

a) The magnitude of linear density, $\left|\lambda \right|=3.6\mathrm{nC}/\mathrm{m}$

b) The wire has fixed negative charge as -q .

c) The radius of the cylindrical shell, r=1.5 cm

d) The net external field is zero.

## Step 2: Understanding the concept of Gauss law-planar symmetry

Using the concept of the electric field of a Gaussian cylindrical surface, we can get the value of the linear density of the cylinder for the net external field to be zero. Now, using this linear density, we can get the surface density of the cylinder.

Formulae:

The magnitude of the electric field of a Gaussian cylindrical surface,$\left|E\right|=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_{0}\mathrm{r}}$ (1)

The surface density of a cylinder, $\sigma =\frac{\lambda }{2\mathrm{\pi r}}$

## Step 3: Calculation of the surface charge density

The net electric field for r > R is given by using equation (1) as follows:

${E}_{net}={E}_{wire}+{E}_{cyclinder}\phantom{\rule{0ex}{0ex}}=\frac{-\lambda }{2\mathrm{\pi }{\epsilon }_{0}r}+\frac{{\lambda }^{\text{'}}}{2\mathrm{\pi }{\epsilon }_{0}r}$

But, for the given net external field value to be zero, we get the linear charge density of the cylinder from the above equation as:${\lambda }^{\text{'}}=3.6\mathrm{nC}/\mathrm{m}$

Now, substituting this value in equation (2), we can get the value of the surface charge density as follows:

$\mathrm{\sigma }=\frac{3.6×{10}^{-6}\mathrm{C}/\mathrm{m}}{\left(2\mathrm{\pi }\right)\left(0.015\mathrm{m}\right)}\phantom{\rule{0ex}{0ex}}=3.8×{10}^{-8}\mathrm{C}/\mathrm{m}$

Hence, the value of the surface charge density is $3.8×{10}^{-8}\mathrm{C}/{\mathrm{m}}^{2}$ .

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