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REFERENCE: https://drive.google.com/file/d/1DXujtLSLCiQLeo0RhJnFaPVTLnh4A1_u/view?usp=sharing REFER TO THIS LINK: https://drive.google.com/file/d/1DXujtLSLCiQLeo0RhJnFaPVTLnh4A1_u/view?usp=sharing please answer this, with solutions and explanations, what's more and what i have learned. I WILL GIVE A
REFERENCE: https://drive.google.com/file/d/1DXujtLSLCiQLeo0RhJnFaPVTLnh4A1_u/view?usp=sharing
REFER TO THIS LINK: https://drive.google.com/file/d/1DXujtLSLCiQLeo0RhJnFaPVTLnh4A1_u/view?usp=sharing
please answer this, with solutions and explanations, what's more and what i have learned. I WILL GIVE A HIGH RATE. thank you so much. please help.
Ito Ang GAUSS-to Kong Problemahin Il. Problem-Solving: Solve the following problems and follow the format: Given, Illustration, Asked, and Solution. Then, enclose in a box the final answer. Write your answer on a separate sheet of paper. 1. A uniform electric field of magnitude 5.8 x 10 N/C passes through a circle of radius 13 cm. What is the electric flux through the circle when its face is (a) perpendicular to the field lines, (b) at 45" to the field lines, and (c) parallel to the field lines? 2. In the figure below, two objects O, and Oz, have charges +1 pC and -2 pC respectively, and a third object, Os, is electrically neutral. (a) what is the electric flux through the surface A, that encloses all the three objects? (b) what is the electric flux through the surface Az that encloses the third object only? Al 01 02 3. A 15-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The total electric flux leaving the bag is 7.3 x 10" Nm /C. What is the linear charge density (A) on the rod? 4. The following charges are located inside a submarine: 4.50 pC, -9.00 pC, 27.0 pC, and - 88.0 pC. Calculate the net electric flux through the hull of the submarine. 5. Consider an infinitely long, very thin metal tube with radius R=2.90 cm. If the linear charge density of the cylinder is A= 1.50 x 10C/m, what is the approximate magnitude of the electric field at radial distance r=2R? What I Have Learned Instructions: Supply the missing termis in each item to complete the thought of the sentence. 1. is a measure of the number of field lines passing perpendicularly through a surface. 2. states that the total electric flux through a surface is the total electric charge (qual) inside the surface divided by En. 3. is known as a closed surface in three-dimensional space such that the flux vector field is calculated. 4. When surface is to the electric fields it means no electric field lines cross the surface. 5. For a the charges reside at its surface area.What's More It's Your Turn! L. INTERPRETIVE TYPE. A. On your answer sheet, draw and color the @ if you agree with the statement. Draw and color the @ if you disagree with the statement. EMOTICON STATEMENT 1. Electric flux (DE) is a measure of the number of field lines passing parallel through a surface. 2. Electric flux is a vector quantity, which means it has magnitude and direction. 3. The flux through a solid surface is positive if the field is directed out of the region contained by the surface. 4. The charge distribution may be expressed in terms of linear charge density, surface charge density, or volume charge density. 5. You don't need to identify a Gaussian body; you need a Gaussian surface. B. Draw and color the (@ if the second clause agrees with the first clause. Draw and color () if it does not. First Clause Conjunction Second Clause Emoticon 1. The electric flux is DE = 0 because the electric field (E) is parallel (edge-on) to the surface. 2. It is a Gaussian surface because the flux or electric field is of a cylinder produced on a point charge. 3. The magnitude of the because the charges of a conductor electric field inside a reside at its surface. conducting sphere is equal to zero 4. The total flux through the | because the negative flux directed entire closed surface is into the region is canceled zero out by the positive flux directed out of the region. 5. If the electric field makes | because the magnitude of the flux an angle with the direction should be proportional to the normal to the flat surface, it component of the field will be solved using the perpendicular to the surface. formula DE = EA cos 0Step by Step Solution
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