Concept explainers
In a certain region of space, the electric field is
(a)
The volume density of electric charge at
Answer to Problem 24.42P
The volume density of electric charge at
Explanation of Solution
Given info: The electric field vector in the region of space is
Consider a Gaussian box of thickness
Formula to calculate the electric flux at
Here,
Write the expression for the electric field at
Substitute
Thus, electric field in the region of space at
The electric field enter into the box at
Substitute
Formula to calculate the electric flux at
Here,
Write the expression for the electric field at
The term
The electric field enter into the box at
Substitute
Write the expression for the net electric flux through the box.
Substitute
Write an alternate expression for the net electric flux from Gauss law.
Here,
Formula to calculate the volume of the Gaussian box is,
Here,
Formula to calculate the average charge enclosed inside the rectangular Gaussian surface is,
Here,
Replace
Substitute
Equate the equation (2) and (1) for same value of electric flux.
Substitute
Conclusion:
Therefore, the volume density of electric charge at
(b)
Whether the given region of space can be inside a conductor.
Answer to Problem 24.42P
No, the given region of space could not be inside a conductor.
Explanation of Solution
Given info: The electric field vector in the region of space is
According to the principle of Electromagnetism, in electrostatics free charges in a good conductor reside only on the surface. So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is uniform, the electric fields cancel each other. No matter, what is the shape of the conductor as long as there is field inside it, electrons always rearrange themselves to make the net field zero.
But, the volume density of electric charge at
Conclusion:
Therefore, the given region of space could not be inside a conductor.
Want to see more full solutions like this?
Chapter 24 Solutions
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
- air is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cmarrow_forwardNo chatgpt pls will upvotearrow_forward13.87 ... Interplanetary Navigation. The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit (Fig. P13.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the depar- ture planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. (a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth? (b) How long does a one- way trip from the the earth to Mars take, between the firings of the rockets? (c) To reach Mars from the…arrow_forward
- No chatgpt pls will upvotearrow_forwarda cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?arrow_forwardCalculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were: 222.22 800.00 61.11 641.67 0.00 588.89 11.11 588.89 8.33 588.89 11.11 588.89 5.56 586.11 2.78 583.33 Give in the answer window the calculated repeated experiment variance in m/s2.arrow_forward
- How can i solve this if n1 (refractive index of gas) and n2 (refractive index of plastic) is not known. And the brewsters angle isn't knownarrow_forward2. Consider the situation described in problem 1 where light emerges horizontally from ground level. Take k = 0.0020 m' and no = 1.0001 and find at which horizontal distance, x, the ray reaches a height of y = 1.5 m.arrow_forward2-3. Consider the situation of the reflection of a pulse at the interface of two string described in the previous problem. In addition to the net disturbances being equal at the junction, the slope of the net disturbances must also be equal at the junction at all times. Given that p1 = 4.0 g/m, H2 = 9.0 g/m and Aj = 0.50 cm find 2. A, (Answer: -0.10 cm) and 3. Ay. (Answer: 0.40 cm)please I need to show all work step by step problems 2 and 3arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College