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Chapter 24: Q. 48 (page 685)

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.

Short Answer

Expert verified

(part a) The electric field strengths E1to E5in regions 1to 5.

a)

E1=3Q2A系0E2=0E3=Q2A系0E4=0E5=3Q2A系0

(part b) The surface charge densities 畏uto 畏don the four surface a to d.

b)

畏a=+3Q2A畏b=-Q2A畏c=+Q2A畏d=+3Q2A

Step by step solution

01

Step 1:

dff

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Most popular questions from this chapter

The conducting box in FIGURE EX24.26 has been given an excess negative charge. The surface density of excess electrons at the center of the top surface is 5.0脳1010electrons/m2 . What are the electric field strengths E1to E3at points 1 to 3?

A 1.0cm脳1.0cm脳1.0cm box with its edges aligned with the xyz-axes is in the electric field E鈫�=(350x+150)i^N/C, where x is in meters. What is the net electric flux through the box?

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鈮�a,a<r<b,b鈮�r鈮�c,andr>c.

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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?
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