Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Question
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Chapter 4, Problem 41P

 (a)

To determine

 The total charge.

 (a)

Expert Solution
Check Mark

Explanation of Solution

 Given:

 The atomic radius is a.

 The volume charge density (ρv) is ρo(1r2a2).

 Calculation:

 Calculate the total charge (Q) using the relation.

Q=vρvdvQ=ϕ=02πθ=0πr=0aρvr2sinθdrdθdϕ

Q=ϕ=02πθ=0πr=0aρo(1r2a2)r2sinθdrdθdϕ=ρoa2ϕ=02πθ=0πr=0a(r2r4)sinθdrdθdϕ=ρoa2[r33r55]0a[cosθ]0π[ϕ]02π=ρoa2[a33a550][1+1][2π0]=4πρo(a3a35)

 Thus, the total charge (Q) is 4πρo(a3a35)_.

 (b)

To determine

 The field E and potential V outside the nucleus.

 (b)

Expert Solution
Check Mark

Explanation of Solution

 Given:

 The atomic radius is a.

 The volume charge density (ρv) is ρo(1r2a2).

 Calculation:

 Outside the nucleus, the radius is r>a.

 Calculate the electric field intensity (E) using the relation.

E=Q4πε0r2ar

E=4πρo(a3a35)4πε0r2=ρo(a3a35)ε0r2

 Calculate the potential (V) using the relation.

V=rEdr

V=(ρo(a3a35)ε0r2)dr=ρo(a3a35)ε01r2dr=ρo(a3a35)ε0(1r)=ρo(a35a3)ε0r

 Thus, the electric field intensity (E) is ρo(a3a35)ε0r2_ and the potential is ρo(a35a3)ε0r_.

 (c)

To determine

 The field E and potential V inside the nucleus.

 (c)

Expert Solution
Check Mark

Explanation of Solution

 Given:

 The atomic radius is a.

 The volume charge density (ρv) is ρo(1r2a2).

 Calculation:

 Inside the nucleus, the radius is r<a.

 Calculate the electric field intensity (E) using the relation.

E=Q4πε0r2arE=4πρo(a3a35)4πε0r2

 Here, r<a. Thus, ra and ar.

E=ρoε0a2(r3r35)

 Calculate the potential (V) using the relation.

V=rEdr

V=(ρo(r3r35)ε0a2)dr=ρoε0a20r(r3r35)dr=ρoε0a2(r26r420)

 Thus, the electric field intensity (E) is ρoε0a2(r3r35)_ and the potential is ρoε0a2(r26r420)_.

 (d)

To determine

 The prove that E is maximum at r=0.745a.

 (d)

Expert Solution
Check Mark

Explanation of Solution

 Given:

 The atomic radius is a.

 The volume charge density (ρv) is ρo(1r2a2).

 Calculation:

 Calculate the radius (r) for the maximum electric field using the relation.

dEdr=0

ddr(ρo(r3r35)ε0a2)=0ρoε0a2[ddr(r3r35)]=0ρoε0a2(133r25)=013=3r25

r=0.745

 Thus, the electric field intensity (E) is maximum at r=0.745 not at r=0.745a.

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Chapter 4 Solutions

Elements Of Electromagnetics

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