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Chapter 24: Q. 25 (page 684)

The electric field strength just above one face of a copper penny is 2000N/C. What is the surface charge density on this face of the penny?

Short Answer

Expert verified

The surface charge density on this face of the penny is17.1x10-9C/m²

Step by step solution

01

Given information and Theory used 

Given :

Electric field strength : 2000N/C

Theory used :

The electric field inside a conductor is zero at all times when it is in electrostatic equilibrium. However, all surplus charges on the conductor accumulate on the outside surface, and as further charges are added, they spread out on the outer surface until they reach the electrostatic equilibrium points.

The electric field at the surface of a charged conductor is given by the equationrole="math" localid="1649667892464" Esurface=ηε0 (1)

whereη is the surface charge density, which is a physical parameter that relies on the conductor's form.

02

Calculating the surface charge density on this face of the penny.

We will calculate the surface charge density using the electric field outside the conductor. Rearranging equation (1) for η, we get :

η=Esurface·ε0 (2)

We now input the values for into equation (2) to get

η=Esurface·ε0=(2000N/C)(8.85x10-12Nm²/C²)=17.1x10-9C/m²

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Most popular questions from this chapter

aA uniformly charged ball of radiusaand charge -Qis at the center of a hollow metal shell with inner radius band outer radius c. The hollow sphere has net charge+2Q. Determine the electric field strength in the four regionsr≤a,a<r<b,b≤r≤c,andr>c.

FIGURE EX24.2 shows a cross section of two concentric spheres. The inner sphere has a negative charge. The outer sphere has a positive charge larger in magnitude than the charge on the inner sphere. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

The electric field is constant over each face of the cube shown in FIGURE EX24.4. Does the box contain positive charge, negative charge, or no charge? Explain.

What is the net electric flux through the torus (i.e., doughnut shape) of FIGURE ?

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/ϵ1, where ϵis the permittivity of the material. (Technically, ϵ0is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density 250nC/mis covered with insulation whose permittivity is 2.5ϵ0. What is the electric field strength at a point inside the insulation that is 1.5mmfrom the axis of the wire?
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