Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 5, Problem 40P
To determine

Calculate the value of D2 and the angle θ2 that D2 makes with the normal.

Expert Solution & Answer
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Answer to Problem 40P

The value of D2 is 5.96ax9.28ay+16.25aznC/m2 and the angle θ2 that D2 makes with the normal is 87.66°.

Explanation of Solution

Calculation:

Given, the electric interface is defined by 4x+3y=10m and the electric flux density is

D1=2ax4ay+6.5aznC/m2

Find the unit vector normal to the interface as follows,

an=4ax+3ay42+32=4ax+3ay5=0.8ax+0.6ay

Consider the general expression for dielectric-dielectric interface.

D1n=D2n        (1)

Find the value D1n.

D1n=D1an

Substitute (2ax4ay+6.5az)×109 for D1 and 0.8ax+0.6ay for an in above expression.

D1n=(2ax4ay+6.5az)×109[(0.8ax+0.6ay)]=0.8nC/m2

Therefore,

D1n=D1nan

Substitute (0.8)×109 for D1n and 0.8ax+0.6ay for an in above expression

D1n=(0.8)×109(0.8ax+0.6ay)=(0.64ax0.48ay)×109=0.64ax0.48aynC/m2

Therefore, from Equation (1),

D1n=D2n=0.64ax0.48aynC/m2

Find the value of D1t.

D1t=D1D1n

Substitute (2ax4ay+6.5az)×109 for D1 and (0.64ax0.48ay)×109 for D1n in above equation.

D1t=[(2ax4ay+6.5az)×109][(0.64ax0.48ay)×109]=(2.64ax3.52ay+6.5az)×109=(2.64ax3.52ay+6.5az)nC/m2

Consider the general expression for dielectric-dielectric interface.

E1t=E2t

D1tεr1=D2tεr2        (2)

As region including the origin in free space, εr1=1.

Substitute 1 for εr1, 2.5 for εr2, and (2.64ax3.52ay+6.5az)×109 for D1t in Equation (2).

(2.64ax3.52ay+6.5az)×1091=D2t2.5D2t=2.5(2.64ax3.52ay+6.5az)×109D2t=6.6ax8.8ay+16.25aznC/m2

Find the value of D2.

D2=D2t+D2n

Substitute (6.6ax8.8ay+16.25az)×109 for D2t and (0.64ax0.48ay)×109 for D2n in above expression.

D2=[(6.6ax8.8ay+16.25az)×109]+[(0.64ax0.48ay)×109]=(5.96ax9.28ay+16.25az)×109=5.96ax9.28ay+16.25aznC/m2

Consider the general expression to find the angle θ2 that D2 makes with the normal.

tanθ2=|D2t||D2n|=6.62+8.82+16.2520.642+0.482=24.53θ2=tan1(24.53)

The above equation becomes,

θ2=87.66°

Conclusion:

Thus, the value of D2 is 5.96ax9.28ay+16.25aznC/m2 and the angle θ2 that D2 makes with the normal is 87.66°.

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