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Chapter 4: Q22-35P (page 747)

22.35: The electric filed \(\overrightarrow E \) in Fig. 22.35 is everywhere parallel to the \(x\)-axis, so the components \({E_y}\) and \({E_z}\) are zero. The \(x\) component of the field \({E_x}\) depends on \(x\) but not on \(y\)or \(z\). At points in the \(yz\) plane (where \(x = 0\)), \({E_x} = 125\,N/C\). (a) what is the electric flux through surface \(I\) in Fig 22.35? (b) what is the electric flux through the surface \(II\)? (c) The volume shown is a small section of a very large insulating slab \(1.0\,m\) thick. If there is a total charge of \( - 24.0\,nC\) within the volume shown, what are the magnitude and direction of \(\overrightarrow E \) at the face opposite surface \(I\)? (d) Is the electric field produced by charges only within the slab, or is the field also due to charges outside the slab? How can you tell?

Short Answer

Expert verified
  1. The electric flux through the surface \(I\) is \(750\,N{m^2}/C\).

Step by step solution

01

Identification of given data

The electric field in the\(x\)direction is given as\({E_x} = 125\,N/C\).

02

Understanding the concept:

The expression for the electric flux is given by,

\(\phi = \overrightarrow E \cdot \overrightarrow A \)鈥︹�� (1)

Here\(\phi \)is the electric flux,\(\overrightarrow E \)is the electric field vector,\(\overrightarrow A \)is the area vector.

Note that electric flux \(\phi \) is scalar quantity, because dot product of two vector quantity is scalar.

03

Determine the electric flux through first surface:

The value of the area vector is,

\(\begin{aligned}\overrightarrow A = \left( {2 \times 3} \right)\mathop i\limits^ \wedge \\ = 6\mathop i\limits^ \wedge \end{aligned}\)

At\(x = 0\),\(\overrightarrow {\left| {{E_x}} \right|} = 125\,N/C\),\({E_y} = 0\)

So,\(\overrightarrow E = 125\,\mathop i\limits^ \wedge \)

Substitute\(125\,\mathop i\limits^ \wedge \)for\(\overrightarrow E \),\(6\mathop i\limits^ \wedge \)for\(\overrightarrow A \)into the equation (1),

\(\begin{aligned}{\phi _1} = \left( {125\,\mathop i\limits^ \wedge } \right) \cdot \left( {6\mathop i\limits^ \wedge } \right)\\ = 750\,N{m^2}/C\end{aligned}\)

Therefore the electric flux through the surface \(I\) is \(750\,N{m^2}/C\).

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Most popular questions from this chapter

A horizontal rectangular surface has dimensions 2.80cmby 3.20cmand is in a uniform magnetic field that is directed at an angle of 30.0掳above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10脳10-4Wb through the surface?

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Fig. E25.29.
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