Implementing Gauss's law on a cylinder.You're going to find the electric field at a point P due to a cylinder made of an insulating material.Again, we'll assume an infinitely long cylinder, although you can consider a section with finite lengthL, and radius R. Let P be a distance "D" away from the central axis of the cylinder, which has a chargedensity p.Make sure to keep D and R straight, in terms of which you are using where. The drawing can help.1. Draw a suitable Gaussian surface on the picture below.PL2. Add vectors for då of the gaussian surface, (the infinitesimal area elements) on the drawing.3. Draw vectors for the electric field caused by L on the Gaussian surface.4. Write down an equation for p. Plug in to a geometrical equation so that it is in terms of givenvariables.5. Write down Gauss's law, and use # 4 to substitute on one side.6. Simplify the other side of Gauss's law based on the geometry of what you did in #2 and #3.7. Use an equation for the surface area of the shape you drew to eliminate the integral, then solve forthe electric field.

A insulating cylinder infinitely long with radius R. let P be of a distance D away from central axis…

Implementing Gauss's law on a cylinder. You're going to find the electric field at a point P due to a cylinder made of an insulating material. Again, we'll assume an infinitely long cylinder, although you can consider a section with finite length L, and radius R. Let P be a distance "D" away from the central axis of the cylinder, which has a charge density p. Make sure to keep D and R straight, in terms of which you are using where. The drawing can help. 1. Draw a suitable Gaussian surface on the picture below. P 2. Add vectors for då of the gaussian surface, (the infinitesimal area elements) on the drawing. 3. Draw vectors for the electric field caused by L on the Gaussian surface. 4. Write down an equation for p. Plug in to a geometrical equation so that it is in terms of given variables. 5. Write down Gauss's law, and use #4 to substitute on one side. 6. Simplify the other side of Gauss's law based on the geometry of what you did in #2 and # 3. 7. Use an equation for the surface area of the shape you drew to eliminate the integral, then solve for the electric field.

Implementing Gauss's law on a cylinder. You're going to find the electric field at a point P due to a cylinder made of an insulating material. Again, we'll assume an infinitely long cylinder, although you can consider a section with finite length L, and radius R. Let P be a distance "D" away from the central axis of the cylinder, which has a charge density p. Make sure to keep D and R straight, in terms of which you are using where. The drawing can help. 1. Draw a suitable Gaussian surface on the picture below. P 2. Add vectors for då of the gaussian surface, (the infinitesimal area elements) on the drawing. 3. Draw vectors for the electric field caused by L on the Gaussian surface. 4. Write down an equation for p. Plug in to a geometrical equation so that it is in terms of given variables. 5. Write down Gauss's law, and use #4 to substitute on one side. 6. Simplify the other side of Gauss's law based on the geometry of what you did in #2 and # 3. 7. Use an equation for the surface area of the shape you drew to eliminate the integral, then solve for the electric field.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter25: Gauss’s Law

Section: Chapter Questions

Problem 68PQ: Examine the summary on page 780. Why are conductors and charged sources with linear symmetry,...

See similar textbooks

Similar questions

As shown in Figure, a particle of charge +Q produces an electric field of magnitude Epart at point P, atdistance R from the particle. As in figure b, that same amount of charge is spread uniformly along a circulararc that has radius R and subtends an angle ?. The charge on the arc produces an electric field of magnitudeEarc at its center of curvature P. For what value of ? does Earc = 0.50 Epart ?
(FILL IN THE BLANKS) Problem Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and total charge Q) at x0 distance from its center? (Consider that the surface of the plate lies in the yz plane)
Examine the summary on page 780. Why are conductors and charged sources with linear symmetry, spherical symmetry, and planar symmetry categorized as special cases rather than major concepts or underlying principles?
Reformulate Gauss’s law by choosing the unit normal of the Gaussian surface to be the one directed inward.
Consider the three-dimensional conductor in the figure, that has a hole in the center. The conductor has an excess charge 7.2 μC on it. What is the electric flux (in N⋅m2/C) through the Gaussian surface S1 shown in the figure? Now put a point of charge 27.4 μC inside the cavity of the conductor. What is the flux (in N⋅m2/C) through the Gaussian surface S1? Now consider the Gaussian surface S2. With the charge still inside the cavity, what is the flux (in Nm2/C) through this surface?
The charge per unit length on the thin rod shown below is λ . What is the electric field at the point P? (Hint: Solve this problem by first considering the electric field dE at P due to a small segment dx of the rod, which contains charge dq = λdx . Then find the net field by integrating dE over the length of the rod.)
Find the electric field at the center of a hemispherical shell of charge Q and radius R.
Figure 2 shows a nonconducting rod with a uniformly distributed charge Q. The rod forms a half-circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P, by what factor is the magnitude of the electric field at P multiplied?
Two concentric spherical shells with radius “R1” and “R2” , charge quantity “Q1” and “Q2” respectively, find the distribution of electric field strength in space , and make the E-r relationship curve.
Find the electric field everywhere of an infinite uniform line charge with total charge Q. Sol. Using Gauss's Law: (contour integral E with rightwards harpoon with barb upwards on top) * (stack d A with rightwards harpoon with barb upwards on top) = ___________ /epsilon0 EA = ___________ / epsilon0 Since, its the line charge we use the area of a cylinder surrounding the line charge E*2pi ___________ L= ___________ / epsilon0 But all the charged get enclosed by the cylinder area, so qenc = Q Deriving, we get: E = (1/2(pi*epsilon0))*(___________ /rL) **see picture for the symbols used
Find the electric field at a distance h from the center of a ring of radius R, whose linear charge density is homogeneous and equal to λ
Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at x=0 has positive charge density η, and the disk centered at x=a has negative charge density −η, where the charge density is charge per unit area. What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2? Express your answer in terms of η, R, a, and the permittivity of free space ϵ0.