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Q15P

Expert-verifiedFound in: Page 680

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A particle of charge **${\mathbf{+}}{\mathit{q}}$ **is placed at one corner of a Gaussian cube. What multiple of **${\mathit{q}}{\mathbf{/}}{{\mathit{\epsilon}}}_{{\mathbf{0}}}$ **gives the flux through (a) each cube face forming that corner and (b) each of the other cubes faces?**

- The multiple that gives the flux through each cube face is zero.
- The multiple that gives the flux through each of the cube’s faces are 1/24 or 0.0417.

The total flux through any surface that completely surrounds the point charge is $q/{\epsilon}_{0}$

**Using the concept of Gauss law and the planar symmetry, we can get the net flux through each cube face by considering the concept of total flux through the cube.**

If we stack identical cubes side by side and directly on top of each other, we will find that eight cubes meet at any corner.

Thus, one-eighth of the field lines emanating from the point charge pass through a cube with a corner at the charge, and the total flux through the surface of such a cube is $q/8{\epsilon}_{0}$.

Now the field lines are radial, so at each of the three cube faces that meet at the charge, the lines are parallel to the face and the flux through the face is zero.

The fluxes through each of the other three faces are the same, so the flux through each of them is one-third of the total.

That is, the flux through each of these faces is given as:

$\frac{1}{3}\times \left(\frac{\mathrm{q}}{8{\mathrm{\epsilon}}_{0}}\right)=\frac{\mathrm{q}}{24{\mathrm{\epsilon}}_{0}}$

Thus, the multiple factors are 1/24 or 0.0417.

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