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A point charge of +5.00 C is located on the x-axis at x =
Chapter 22, Problem 13E(choose chapter or problem)
A point charge of +5.00 ???C is located on the ?x?-axis at ?x = 4.00 m, next to a spherical surface of radius 3.00 m centered of the origin. (a) Calculate the magnitude of the electric field at ?x = 3.00 m. (b) Calculate the magnitude of the electric held at ?x = 3.00 m. (c) According to Gauss’s law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e at ?x = 3.00 m) than on the far side (at ?x = – 3.00 m). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.
Questions & Answers
QUESTION:
A point charge of +5.00 ???C is located on the ?x?-axis at ?x = 4.00 m, next to a spherical surface of radius 3.00 m centered of the origin. (a) Calculate the magnitude of the electric field at ?x = 3.00 m. (b) Calculate the magnitude of the electric held at ?x = 3.00 m. (c) According to Gauss’s law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e at ?x = 3.00 m) than on the far side (at ?x = – 3.00 m). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.
ANSWER:Solution 13E Step 1 : In this question, we need to find the electric field at x=3.0 m In the second part, we need to find magnitude of electric field at x=-3 .0 m In the third part, using Gauss’s law , we need to explain how the flux into the sphere is equal to the flux out of it, with the help of a sketch Data given Q = 5 × 10 C6 Distance on x-axis d = 4.0 m