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Q12Q

Expert-verifiedFound in: Page 678

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure 23-29 shows four Gaussian surfaces consisting of identical cylindrical midsections but different end caps. The surfaces are in a uniform electric field**** $\overrightarrow{E}$**** that is directed parallel to the central axis of each cylindrical midsection. The end caps have these shapes:**** ${{\mathit{S}}}_{{\mathbf{1}}}$****, convex hemispheres;**** ${{\mathit{S}}}_{{\mathbf{3}}}$****, concave hemispheres;**** ${\mathit{S}}_{\mathbf{3}}$****, cones;**** ${{\mathit{S}}}_{{\mathbf{4}}}$****, flat disks. Rank the surfaces according to (a) the net electric flux through them and (b) the electric flux through the top end caps, greatest first.**

a.The net flux through all the surfaces becomes zero.

b.The rank of the surfaces according to the electric flux through top and caps is${S}_{1}={S}_{2}={S}_{3}={S}_{4}$ .

- The figure with four Gaussian surfaces consisting of identical midsections and different end caps are given.
- The surface is uniform to the electric field.
- End caps have shapes ${S}_{1}$, convex hemisphere, ${S}_{2}$ concave hemisphere, ${S}_{3}$ cone, ${S}_{4}$ flat disk.

**Electric flux is the property of an electric field that describes the number of electric field lines crossing through a given surface area. It is also given by the net charge enclosed by the Gaussian surface divided by the permeability of vacuum space. For a given constant electric field, the net flux depends on the surface area of the given Gaussian shapes.**

Formula:

The electric flux through any closed surface,

${\varphi}_{E}=\oint \overrightarrow{E}\cdot \overrightarrow{dA}$ ….. (i)

Here, is the area vector and $\overrightarrow{E}$ is the electric field.

Gauss’ law defined as the flux through any closed surface is $1/{\epsilon}_{0}$ time of the charge enclosed with the surface.

${\varphi}_{E}=\frac{{q}_{enclosed}}{{\epsilon}_{0}}$

Here, ${\epsilon}_{0}$ is the permittivity of free space and ${q}_{enclosed}$ is the charge is the charge enclosed with in the surface.

As, the net charge enclosed within the surface is

${q}_{in}=0$

Thus, the value of net flux for uniform field for any closed surface will be given using equation (i) as:

$\begin{array}{c}{\varphi}_{E}=\oint \overrightarrow{E}.\overrightarrow{dA}\\ =0\end{array}$

The electric field for the cylindrical surfaces points perpendicular to the area vector. Therefore the cylindrical flux is zero. The electric field for the end surface is at zero and with the area vector along this surface. Therefore the flux through these surfaces cancels.

Hence, the net flux through all the surfaces becomes zero.

By comparing equation (1) and (2), you get

$\oint \overrightarrow{E}\cdot \overrightarrow{dA}=\frac{{q}_{enclosed}}{{\epsilon}_{0}}$

In this case, you are having four different surfaces and all closed surfaces.

The direction of the net flux from both the sides of the given surfaces will be outwards.

That means the amount of flux entering the given end cap will be equal to the amount of flux entering through the top cap of the given surfaces ${S}_{1},{S}_{2},{S}_{3},{S}_{4}$ as the flux from the plane area and the curved surface area are same for all the given surfaces that is:

${\varphi}_{Plane}={\varphi}_{Curved}=\text{constant}$

Thus, the rank of the surfaces entering through the top and end caps will remain the same that is ${S}_{1}={S}_{2}={S}_{3}={S}_{4}$.

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