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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use Gauss’s Law to find the charge enclosed by the cube with vertices (±1, ±1, ±1) if the electric field is E(x, y, z) = x i + y j + z k

To determine

To find: The charge enclosed by the cube with vertices (±1,±1,±1) .

Explanation

Given:

E(x,y,z)=xi+yj+zk and cube with vertices (±1,±1,±1) .

Formula used:

From Gauss law,

Q=ε0SEdS (1)

SFdS=DF(ru×rv)dA (2)

rr=xri+yrj+zrk (3)

Here S consists of six faces (S1,S2,S3,S4,S5andS6) of the cube with vertices (±1,±1,±1) is shown in Figure 1.

On S1 surface:

Consider E=i+yj+zk .

Find ry .

Modify equation (3).

ry=xyi+yyj+zyk

Substitute 1 for x , y for y and z for z ,

ry=(1)yi+(y)yj+(z)yk=(0)i+(1)j+(0)k=j

Find rz .

Modify equation (3).

rz=xzi+yzj+zzk

Substitute 1 for x , y for y and z for z ,

rz=(1)zi+(y)zj+(z)zk=(0)i+(0)j+(1)k=k

Find ry×rz .

ry×rz=(j)×(k)=|ijk010001|=(10)i(00)j+(00)k=i

Find S1EdS .

Modify equation (2).

S1EdS=DE(ry×rz)dA

Apply limits and substitute i+yj+zk for E and i for ry×rz ,

S1EdS=1111(i+yj+zk)(1i)dydz=1111(1)dydz=11dy11(1)dz=[y]11[z]11

Simplify the equation.

S1EdS=[1+1][1+1]=(2)(2)=4

On S2 surface:

Consider E=xi+j+zk .

Find rz .

Modify equation (3).

rz=xzi+yzj+zzk

Substitute x for x , 1 for y and z for z ,

rz=(x)zi+(1)zj+(z)zk=(0)i+(0)j+(1)k=k

Find rx .

Modify equation (3).

rx=xxi+yxj+zxk

Substitute x for x , 1 for y and z for z ,

rx=(x)xi+(1)xj+(z)xk=(1)i+(0)j+(0)k=i

Find rz×rx .

rz×rx=(k)×(i)=|ijk001100|=(00)i(01)j+(00)k=j

Find S2EdS .

Modify equation (2).

S2EdS=DE(rz×rx)dA

Apply limits and substitute xi+j+zk for E and j for rz×rx ,

S2EdS=1111(xi+j+zk)(1j)dxdz=1111(1)dxdz=11dx11(1)dz=[x]11[z]11

Simplify the equation.

S2EdS=[1+1][1+1]=(2)(2)=4

On S3 surface:

Consider E=xi+yj+k .

Find rx .

Modify equation (3).

rx=xxi+yxj+zxk

Substitute x for x , y for y and 1 for z ,

rx=(x)xi+(y)xj+(1)xk=(1)i+(0)j+(0)k=i

Find ry .

Modify equation (3).

ry=xyi+yyj+zyk

Substitute x for x , y for y and 1 for z ,

ry=(x)yi+(y)yj+(1)yk=(0)i+(1)j+(0)k=j

Find rx×ry .

rx×ry=(i)×(j)=|ijk100010|=(00)i(00)j+(10)k=k

Find S3EdS .

Modify equation (2).

S3EdS=DE(rx×ry)dA

Apply limits and substitute xi+yj+k for E and k for rx×ry ,

S3EdS=1111(xi+yj+k)(1k)dxdy=1111(1)dxdy=11dx11(1)dy=[x]11[y]11

Simplify the equation.

S3EdS=[1+1][1+1]=(2)(2)=4

On S4 surface:

Consider E=i+yj+zk .

Find rz .

Modify equation (3).

rz=xzi+yzj+zzk

Substitute 1 for x , y for y and z for z ,

rz=(1)zi+(y)zj+(z)zk=(0)i+(1)j+(1)k=k

Find ry

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Chapter 16 Solutions

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Sect-16.1 P-11ESect-16.1 P-12ESect-16.1 P-13ESect-16.1 P-14ESect-16.1 P-15ESect-16.1 P-16ESect-16.1 P-17ESect-16.1 P-18ESect-16.1 P-21ESect-16.1 P-22ESect-16.1 P-23ESect-16.1 P-24ESect-16.1 P-25ESect-16.1 P-26ESect-16.1 P-29ESect-16.1 P-30ESect-16.1 P-31ESect-16.1 P-32ESect-16.1 P-33ESect-16.1 P-34ESect-16.1 P-35ESect-16.1 P-36ESect-16.2 P-1ESect-16.2 P-2ESect-16.2 P-3ESect-16.2 P-4ESect-16.2 P-5ESect-16.2 P-6ESect-16.2 P-7ESect-16.2 P-8ESect-16.2 P-9ESect-16.2 P-10ESect-16.2 P-11ESect-16.2 P-12ESect-16.2 P-13ESect-16.2 P-14ESect-16.2 P-15ESect-16.2 P-16ESect-16.2 P-17ESect-16.2 P-18ESect-16.2 P-19ESect-16.2 P-20ESect-16.2 P-21ESect-16.2 P-22ESect-16.2 P-23ESect-16.2 P-24ESect-16.2 P-25ESect-16.2 P-26ESect-16.2 P-31ESect-16.2 P-32ESect-16.2 P-33ESect-16.2 P-34ESect-16.2 P-35ESect-16.2 P-36ESect-16.2 P-37ESect-16.2 P-38ESect-16.2 P-39ESect-16.2 P-40ESect-16.2 P-41ESect-16.2 P-42ESect-16.2 P-43ESect-16.2 P-44ESect-16.2 P-45ESect-16.2 P-46ESect-16.2 P-47ESect-16.2 P-48ESect-16.2 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