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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/m虏

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=畏蔚0 (1)

where畏 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 畏, we get :

畏=Esurface路蔚0 (2)

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

畏=Esurface路蔚0=(2000N/C)(8.85x10-12Nm虏/C虏)=17.1x10-9C/m虏

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Most popular questions from this chapter

What is the electric flux through the surface shown in FIGURE EX24.9?

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.

II An infinite slab of charge of thickness 2z0lies in the XYplane between z=鈭�z0andz=+z0. The volume charge density 蚁C/m3is a constant.

a. Use Gauss's law to find an expression for the electric field strength inside the slab 鈭�z0鈮�z鈮�z0.

b. Find an expression for the electric field strength above the slab z鈮�z0.

c. Draw a graph of Efrom z=0toz=3z0.

A sphere of radius Rhas total charge Q. The volume charge Calc density role="math" localid="1648722354966" Cm3within the sphere is 蚁r=Cr2, whereC is a constant to be determined.
a. The charge within a small volume dVis dq=蚁dV. The integral of 蚁dVover 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鈮�R in terms of QandR.
c. Does your expression have the expected value at the surface,r=R ? Explain.

FIGURE P24.48shows two very large slabs of metal that are parallel and distance lapart. The top and bottom of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1has total charge Q1=Qand metal 2has total charge Q2=2Q. Assume Qis positive. In terms of Qand A, determine

a. The electric field strengths E1toE5in regions 1to 5.

b. The surface charge densities 畏uto 畏don the four surfaces a to d.
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