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Expert-verified Found in: Page 684 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # The net electric flux through each face of a die (singular of dice) has a magnitude in units of ${{\mathbf{10}}}^{{\mathbf{3}}}{\mathbf{}}{{\mathbf{Nm}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{C}}$ that is exactly equal to the number of spots N on the face (1 through 6). The flux is inward for N odd and outward for N even. What is the net charge inside the die?

The net charge inside the die is $2.66×{10}^{-8}\mathrm{C}$ .

See the step by step solution

## Step 1: Listing the given quantities

• The net electric flux = ${10}^{3}\mathrm{N}.{\mathrm{m}}^{2}/\mathrm{C}$

## Step 2: Understanding the concept of Gauss law

Gauss law describes the relation between charge and electric field in a static situation. The equation for Gauss law is,

${{\mathrm{\epsilon }}}_{{0}}{\mathrm{\Phi }}{=}{{\mathrm{q}}}_{{\mathrm{enc}}}$

Here, ${{\mathbit{q}}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$is the net charge inside an imaginary closed surface and is the net flux of the electric field through the surface.

## Step 3: Net charge inside the die

Let ${\varnothing }_{0}={10}^{3}N.{m}^{2}/C$.

The net flux through the entire surface of the dice is given by:

$\mathit{\varnothing }\mathit{}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{-}\mathit{1}}^{\mathit{6}}{\Phi }_{\mathit{n}}\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{-}\mathit{1}}^{\mathit{6}}{\left(-1\right)}^{\mathit{n}}n{\varphi }_{\mathit{0}}\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{\varphi }_{\mathit{0}}\left(-1+2-3+4-5+6\right)\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{3}{\varphi }_{\mathit{0}}$

Thus, the net charge enclosed is:

role="math" localid="1657360360230" $q={\epsilon }_{0}\varphi \phantom{\rule{0ex}{0ex}}=3{\epsilon }_{0}{\varphi }_{0}\phantom{\rule{0ex}{0ex}}=3\left(8.85×{10}^{-12}{C}^{2}/N.{m}^{2}\right)\left({10}^{3}N.{m}^{2}/C\right)\phantom{\rule{0ex}{0ex}}=2.66×{10}^{-8}C$

Therefore, the net charge inside the die is $=2.66×{10}^{-8}C$ . ### Want to see more solutions like these? 