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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 22 / Problem 22.1

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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A flat sheet of paper of area is oriented so that the

Chapter 22, Problem 22.1

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QUESTION:

A flat sheet of paper of area \(0.250\mathrm{\ m}^2\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

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QUESTION:

A flat sheet of paper of area \(0.250\mathrm{\ m}^2\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

ANSWER:

Step 1 of 4

a) The magnitude of the electric flux through the sheet is given by ${\Phi{}}_{E}=$ $\int\limits_{\ }^{\ }E\ cos\varphi{}$ $dA$

Where $\varphi{}$ is the angle between the normal to the sheet $\widehat{n}$ and the electric field $\vec{E}.$

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