Which of the following statements is incorrect? A. Gauss's law expresses the relationship between electric charge...
Which of the following statements is incorrect?
| A. |
Gauss's law expresses the relationship between electric charge and electric field. |
|
| B. |
The electric flux is dependent on the net electric charge inside the surface. |
|
| C. |
Gauss's law can be used to study electric flux of through an irregular surface. |
|
| D. |
The electric flux is dependent on the radius of the sphere. |
|
| E. |
Total electric flux through a closed surface is dependent on the total electric charge inside the surface. |
Homework Answers
The staement(B) is incorrect which is " The electric flux is dependent upon the net electric charge inside the surface."
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Which of the following statements is incorrect?
A.
Gauss's law expresses the relationship between electric charge...
#8 Gauss's Law and The Shell Theorem Consider a hollow sphere with charge uni- formly distributed...
#8 Gauss's Law and The Shell Theorem Consider a hollow sphere with charge uni- formly distributed on its surface. Suppose the total charge is Q, where Q may be positive or negative Recall that Gauss's law as we have seen it is: Qenclosed ΣΕ A = EO where A = 47tr2 is the total area of the Gaussian surface Suppose the sphere radius is Ro and r > Ro. In terms of Gauss's Law, the reason why the electric field...
True or false physics 2 questions. 1. [ ] Gauss's law states that the net electric...
True or false physics 2 questions. 1. [ ] Gauss's law states that the net electric flux ΦE through any closed Gaussian surface is equal to the net charge inside the surface divided by 4πε_0. 2. [ ] Gauss's law is useful for calculating electric field when the charge distribution is highly symmetrical. 3. [ ] At electrostatic equilibrium, the electric field is zero everywhere inside a conductor, and any charge can only be distributed on the surface of the...
Physics 222 Name: Task 7: Electric Flux and Gauss's Law Objective: To illustrate the concept of...
Physics 222 Name: Task 7: Electric Flux and Gauss's Law Objective: To illustrate the concept of "flux" and to avoid performing the integral in Gauss's Law for Electricity 1. A cylindrical Gaussian surface (length = 0.80 m, diameter = 0.20 m) is drawn in an area of a non-uniform electric field, directed at all points in the ex direction (i.e. horizontal to the right). The field varies with x, but does not vary with respect to y or z. The...
Q3: Gauss's Law Problem Statement A-100 nC point charge sits at the center of a hallow...
Q3: Gauss's Law Problem Statement A-100 nC point charge sits at the center of a hallow spherical shell. The shell with radius 0.1 cm and negligible thickness, has a net charge of 200 nC. Find the electric field strength a) inside the sphere at r=0.05cm from the center, and b) outside the sphere at r=0.15cm from the center. In what direction does the electric field point in each case? Visual Representation • Draw a sketch of the charge distribution. •...
QUESTION 6 Gauss's Law describes the following relationship (select the correct answer): O a. Electromotive force...
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2. (10 points) in class we found the Gauss's law for electricity gave us a value...
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2. (10 points) In class we found the Gauss's law for electricity gave us a value...
2. (10 points) In class we found the Gauss's law for electricity gave us a value for the electric field a distance r away (E = (1/41ɛo) (Aenclosed/r?). You have a conductive solid sphere (radius of a) inside a conductive shell (inner radius of b and outer radius of c). The sphere has a charge of +4Q while the shell has a charge of +10Q. Find the following: a. What is the electric field for r<a? What is the electric...
GAUSS LAW PLEASE ? NEED HELP
Which of the following statements about Gauss's law are correct? (There may be more than one correct choice.)Choose all that apply.The electric flux passing through a Gaussian surface depends only on the amount of charge inside that surface, not on its size or shape.Gauss's law is valid only for symmetric charge distributions, such as spheres and cylinders.If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface.Only charge enclosed within...
We find from Gauss's law that the flux through a closed surface: is proportional to the square of the charge enclosed....
We find from Gauss's law that the flux through a closed surface: is proportional to the square of the charge enclosed. is inversely proportional to the volume of the closed surface. is zero. inversely proportional to the net charge enclosed. is proportional to the net charge enclosed.
Gauss's Law in 3, 2, and 1 Dimension
Gauss's Law in 3, 2, and 1 Dimension Gauss's law relates the electric flux \(\Phi_{E}\) through a closed surface to the total charge \(q_{\text {end }}\) enclosed by the surface:Part ADetermine the magnitude \(E(r)\) by applying Gauss's law.Express \(E(r)\) in terms of some or all of the variables/constants \(q, \tau\), and \(\epsilon_{0}\).Part BBy symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of...
#8 Gauss's Law and The Shell Theorem Consider a hollow sphere with charge uni- formly distributed on its surface. Suppose the total charge is Q, where Q may be positive or negative Recall that Gauss's law as we have seen it is: Qenclosed ΣΕ A = EO where A = 47tr2 is the total area of the Gaussian surface Suppose the sphere radius is Ro and r > Ro. In terms of Gauss's Law, the reason why the electric field...
Physics 222 Name: Task 7: Electric Flux and Gauss's Law Objective: To illustrate the concept of "flux" and to avoid performing the integral in Gauss's Law for Electricity 1. A cylindrical Gaussian surface (length = 0.80 m, diameter = 0.20 m) is drawn in an area of a non-uniform electric field, directed at all points in the ex direction (i.e. horizontal to the right). The field varies with x, but does not vary with respect to y or z. The...
Q3: Gauss's Law Problem Statement A-100 nC point charge sits at the center of a hallow spherical shell. The shell with radius 0.1 cm and negligible thickness, has a net charge of 200 nC. Find the electric field strength a) inside the sphere at r=0.05cm from the center, and b) outside the sphere at r=0.15cm from the center. In what direction does the electric field point in each case? Visual Representation • Draw a sketch of the charge distribution. •...
QUESTION 6 Gauss's Law describes the following relationship (select the correct answer): O a. Electromotive force induced in a conductive loop by variation of a magnetic field. b. Voltage difference across a resistance when current is flowing through the resistance. CPhase angles between current and voltage in capacitors and inductors. d. Electric flux through a closed surface and charge enclosed by the surface. QUESTION 7 "Static" electricity can be described as (select the correct answer): a. The buildup of current...
2. (10 points) in class we found the Gauss's law for electricity gave us a value for the electric field a distance r away (E = (1/4xxo) (encloud/r"). You have a conductive solid sphere (radius of a) inside a conductive shell (inner radius of b and outer radius of c). The sphere has a charge of +4Q while the shell has a charge of +100. Find the following: What is the electric field for ra? What is the electric field...
2. (10 points) In class we found the Gauss's law for electricity gave us a value for the electric field a distance r away (E = (1/41ɛo) (Aenclosed/r?). You have a conductive solid sphere (radius of a) inside a conductive shell (inner radius of b and outer radius of c). The sphere has a charge of +4Q while the shell has a charge of +10Q. Find the following: a. What is the electric field for r<a? What is the electric...
We find from Gauss's law that the flux through a closed surface: is proportional to the square of the charge enclosed. is inversely proportional to the volume of the closed surface. is zero. inversely proportional to the net charge enclosed. is proportional to the net charge enclosed.
Gauss's Law in 3, 2, and 1 Dimension Gauss's law relates the electric flux \(\Phi_{E}\) through a closed surface to the total charge \(q_{\text {end }}\) enclosed by the surface:Part ADetermine the magnitude \(E(r)\) by applying Gauss's law.Express \(E(r)\) in terms of some or all of the variables/constants \(q, \tau\), and \(\epsilon_{0}\).Part BBy symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of...
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