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Expert-verified Found in: Page 680 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # An isolated conductor has net charge${\mathbf{+}}{\mathbf{10}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{6}}{\mathbf{}}{\mathbf{C}}$ and a cavity with a particle of charge${\mathbf{=}}{\mathbf{+}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{6}}{\mathbf{}}{\mathbf{C}}$ What is the charge on (a) the cavity wall and (b) the outer surface?

a) The charge on the cavity wall is$-3.0×{10}^{-6}\mathrm{C}$ .

b) The charge on the outer surface is $+1.3×{10}^{-6}\mathrm{C}$ .

See the step by step solution

## Step 1: The given data

a) Net charge on the isolated conductor,${\mathrm{q}}_{\mathrm{iso}}=+10×{10}^{-6}\mathrm{C}$

b) Particle charge in the cavity,$+3.0×{10}^{-6}\mathrm{C}$

## Step 2: Understanding the concept of Gauss law

Using the concept of Gaussian surface, the net charge enclosed in a body is zero and thus, it can be defined by the electric flux passing through the enclosed volume of the surface.

## Step 3: a) Calculation of the charge on the cavity wall

Consider a Gaussian surface that is completely within the conductor and surrounds the cavity. Since the electric field is zero everywhere on the surface, the net charge it encloses is zero. The net charge is the sum of the charge q in the cavity and the charge q(w) on the cavity wall,

Thus, the value of the charge on the cavity wall is given as follows:

$\mathrm{q}+{\mathrm{q}}_{\mathrm{w}}=0\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{\mathrm{w}}=-\mathrm{q}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{\mathrm{w}}=-3.0×{10}^{-6}\mathrm{C}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the charge is$-3.0×{10}^{-6}\mathrm{C}$

## Step 4: b) Calculation of the charge on the outer surface

The net charge Q of the conductor is the sum of the charge on the cavity wall and the charge ${q}_{s}$ on the outer surface of the conductor.

Thus, the value of the charge on the outer surface is given as:

$\mathrm{Q}={\mathrm{q}}_{\mathrm{w}}+{\mathrm{q}}_{\mathrm{s}}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{\mathrm{s}}=\mathrm{Q}-{\mathrm{q}}_{\mathrm{w}}\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{\mathrm{s}}=\left(10×{10}^{-6}\mathrm{C}\right)-\left(-3.0×{10}^{-{6}^{}}\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{q}}_{\mathrm{s}}=+1.3×{10}^{-5}\mathrm{C}$

Hence, the value of the charge is$+1.3×{10}^{-5}\mathrm{C}$ ### Want to see more solutions like these? 