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(a)
The electric flux out of each face.
Answer to Problem 66QAP
The electric flux out of each face is,
Explanation of Solution
Given:
Height of prism,
Depth of prism,
Length of prism,
Electric field,
Formula used:
The electric flux is given by,
Where,
Calculation:
The electric flux is given by,
For surface I:
Thus,
For surface II:
Thus,
For surface III:
Thus,
For surface IV:
Thus,
For surface V:
Thus,
(b)
The net electric flux.
Answer to Problem 66QAP
The net electric flux is,
Explanation of Solution
Height of prism,
Depth of prism,
Length of prism,
Electric field,
Formula used:
The electric flux is given by,
Where,
Calculation:
The net electric flux is,
(c)
The electric field the prism enclosed.
Answer to Problem 66QAP
The electric field enclosed by the prism is,
Explanation of Solution
Given:
Height of prism,
Depth of prism,
Length of prism,
Electric field,
Radius of solid sphere,
Charge,
Formula used:
The electric flux is given by,
Where,
A=Surface area of prism
Calculation:
The electric flux is given by,
Area of the prism is,
Thus,
So, electric field of the prism is,
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Chapter 16 Solutions
COLLEGE PHYSICS,VOLUME 1
- Consider a thin, spherical shell of radius 14.0 cm with a total charge of 32.0 C distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution.arrow_forwardIf more electric field lines leave a gaussian surface than enter it, what can you conclude about the net charge enclosed by that surface?arrow_forwardA solid conducting sphere of radius 2.00 cm has a charge 8.00 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a total charge −4.00 μC. Find the electric field at (a) r = 1.00 cm, (b) r = 3.00 cm, (c) r = 4.50 cm, and (d) r = 7.00 cm from the center of this charge configuration.arrow_forward
- Equation 23-11 (E = s/´0) gives the electric field at points near a charged conducting surface. Apply this equation to a conducting sphere of radius r and charge q, and show that the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.arrow_forwardAt what distance along the central axis of a uniformly charged plastic disk of radius R = 1.31 m is the magnitude of the electric field equal to 1/3 times the magnitude of the field at the center of the surface of the disk?arrow_forwardIn Figure 23-35, short sections of two very long parallel lines of charge are shown, fixed in place and separated by L = 7.6 cm. The uniform linear charge densities are +6.6 µC/m for line 1 and -2.0 µC/m for line 2. Where along the x axis shown is the net electric field from the two lines zero?arrow_forward
- A proton is projected in the positive x direction into a region of a uniform electric fieldarrow_forwardConsider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution.arrow_forward
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning