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Chapter 21: Problem 20

A flat surface with area \(2.0 \mathrm{m}^{2}\) is in a uniform \(850-\mathrm{N} / \mathrm{C}\) electric field. Find the electric flux through the surface when it's (a) at right angles to the field, (b) at \(45^{\circ}\) to the field, and (c) parallel to the field.

Short Answer

Expert verified
The electric flux through the surface is 1700 Nm^2/C when the surface is at right angles to the field, approximately 1201 Nm^2/C when the surface is at a 45 degree angle to the field, and 0 Nm^2/C when the surface is parallel to the field.

Step by step solution

01

Determine the Electric Flux at Right Angles

Since the surface is at right angles to the field, the angle \(\theta\) is 0 degrees. So, we can substitute the given values into our formula for electric flux: \(\Phi_E = EAcos\theta = (850 N/C)(2.0 m^2)cos(0^\circ)\). Using the fact that \(cos(0^\circ) = 1\), we obtain \(\Phi_E = 1700 Nm^2/C\).
02

Determine the Electric Flux at a 45 Degree Angle

Now, the surface is at a 45 degree angle to the field, so \(\theta = 45^\circ\). Substituting these values into our formula: \(\Phi_E = EAcos\theta = (850 N/C)(2.0 m^2)cos(45^\circ)\). Using the fact that \(cos(45^\circ) = \sqrt{2}/2\), we obtain \(\Phi_E = \approx 1201 Nm^2/C\).
03

Determine the Electric Flux When Parallel to the Field

Finally, for the surface parallel to the field, \(\theta = 90^\circ\). So, \(\Phi_E = EAcos\theta = (850 N/C)(2.0 m^2)cos(90^\circ)\). Using the fact that \(cos(90^\circ) = 0\), we get \(\Phi_E = 0 Nm^2/C\). Surfaces parallel to the field do not intersect the field lines, so there is no flux.

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Most popular questions from this chapter

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