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FIGURE EX24.1 shows two cross sections of two infinitely long coaxial cylinders. The inner cylinder has a positive charge, the outer cylinder has an equal negative charge. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

Short Answer

Expert verified

The direction of the electric field for two concentric oppositely charged cylinders is discovered.

Step by step solution

01

Step 1. Given Information

Two cross sections of two infinitely long coaxial cylinders

02

Step 2. Determine that if the inner cylinder has a positive charge and the outer cylinder has an equal negative charge, the electric field diagram shown below.

A positive static charge's electric field always points radially outwards from the charge, while a negative static charge's electric field always points radially inwards towards the charge.

Between two opposed charges, the electric field shifts from positive to negative charge.

Consider two coaxial cylinders with charges of opposing signs. The positive charge is in the inner cylinder, while the negative charge is in the outer cylinder. As a result, the electric field between the cylinders is radially directed from the inner to the outer. The direction of the electric field between the cylinders is depicted in the diagram below:

The following diagram depicts a circular cross-sectional view of the electric field direction between the cylinders:

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Most popular questions from this chapter

A sphere of radius Rhas total charge Q. The volume charge Calc density role="math" localid="1648722354966" Cm3within the sphere is ÒÏr=Cr2, whereC is a constant to be determined.
a. The charge within a small volume dVis dq=ÒÏdV. The integral of ÒÏdVover the entire volume of the sphere is the total chargeQ. Use this fact to determine the constant Cin terms of QandR .
Hint: Let dVbe a spherical shell of radiusr and thicknessdr. What is the volume of such a shell?
b. Use Gauss's law to find an expression for the electric field strengthE inside the sphere, ,r≤R in terms of QandR.
c. Does your expression have the expected value at the surface,r=R ? Explain.

A sphere of radius Rhas total charge Q. The volume charge density C/m3within the sphere is

ÒÏ=ÒÏ01-rR

This charge density decreases linearly from \(\rho_{0}\) at the center to zero at the edge of the sphere.

a. Show that ÒÏ0=3Q/Ï€R3.

b. Show that the electric field inside the sphere points radially outward with magnitude

E=Qr4πϵ0R34-3rR

FIGUREEX24.18shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) -q/ϵ0, (b) q/ϵ0, (c) 3q/ϵ0, and (d) 4q/ϵ0.

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ϵ0, with Qin/ϵ, where ϵ is the permittivity of the material. (Technically,ϵ0 is called the vacuum permittivity.) Suppose that a 50nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is2500N/C . What is the permittivity of rubber?

The electric flux through the surface shown in FIGURE EX24.11 is 25Nm2/C. What is the electric field strength?

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