   Chapter 16.7, Problem 45E

Chapter
Section
Textbook Problem

Use Gauss’s Law to find the charge contained in the solid hemisphere x2 + y2 + z2 ≤ a2, z ≥ 0, if the electric field is E(x, y, z) = x i + y j + 2z k

To determine

To find: The charge contained in the solid hemisphere x2+y2+z2a2 , z0 .

Explanation

Given:

E(x,y,z)=xi+yj+2zk and hemisphere with x2+y2+z2a2 , z0 .

Formula used:

From Gauss law,

Q=ε0SEdS (1)

Sf(x,y,z)dS=Df(r(u,v))(ru×rv)dA (2)

rθ=xθi+yθj+zθk (3)

rϕ=xϕi+yϕj+zϕk (4)

As S consists of two surfaces: S1 is given by z=a2x2y2 and the disk S2 given by 0x2+y2a2,z=0

On S1 surface:

Let x=asinϕcosθ , y=asinϕsinθ , z=acosϕ and limits 0ϕπ2 and 0θ2π .

E=asinϕcosθi+asinϕsinθj+2acosϕk

Find rθ .

Substitute asinϕcosθ for x , asinϕsinθ for y and acosϕ for z in equation (3),

rθ=(asinϕcosθ)θi+(asinϕsinθ)θj+(acosϕ)θk=(acosθ)ϕ(sinϕ)i+(asinθ)ϕ(sinϕ)j+(a)ϕ(cosϕ)k=acosθcosϕi+asinθcosϕjasinϕk

Find rϕ .

Substitute asinϕcosθ for x , asinϕsinθ for y and acosϕ for z in equation (4),

rθ=θ(asinϕcosθ)i+θ(asinϕsinθ)j+θ(acosϕ)k=(asinϕ)θ(cosθ)i+(asinϕ)θ(sinθ)j+(acosϕ)θ(1)k=(asinϕ)(sinθ)i+(asinϕ)(cosθ)j+(acosϕ)(0)k=asinϕsinθi+asinϕcosθj

Find rϕ×rθ .

rϕ×rθ=(acosθcosϕi+asinθcosϕjasinϕk)×(asinϕsinθi+asinϕcosθj)=|ijkacosθcosϕasinθcosϕasinϕasinϕsinθasinϕcosθ0|={(0+a2sin2ϕcosθ)i+(a2sin2ϕsinθ+0)j+(a2cos2θcosϕsinϕ+a2sin2θcosϕsinϕ)k}={a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕ(cos2θ+sin2θ)k}={a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕk} {cos2θ+sin2θ=1}

Find S1EdS .

Modify equation (2) as follows.

S1EdS=DE(r(ϕ,θ))(rϕ×rθ)dA

Apply limits and substitute asinϕcosθi+asinϕsinθj+2acosϕk for E and a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕk for rϕ×rθ ,

S1EdS=02π0π2(asinϕcosθi+asinϕsinθj+2acosϕk)(a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕk)dϕdθ=02π0π2(a3sin3ϕcos2θ+a3sin3ϕsin2θ+2a3cos2ϕsinϕ)dϕdθ=02π0π2(a3sin3ϕ(cos2θ+sin2θ)+2a3cos2ϕsinϕ)dϕdθ=02π0π2(a3sin3ϕ+2a3cos2ϕsinϕ)dϕdθ

Simplify the equation

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