The instantaneous Poynting vector.

Question

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Answer 1

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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I do not understand the process to answer the second part of question b. Please help me understand how to get there!

Rank the six combinations of electric charges on the basis of the electric force acting on 91. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them. ▸ View Available Hint(s) [most negative 91 = +1nC 92 = +1nC 91 = -1nC 93 = +1nC 92- +1nC 93 = +1nC -1nC 92- -1nC 93- -1nC 91= +1nC 92 = +1nC 93=-1nC 91 +1nC 92=-1nC 93=-1nC 91 = +1nC 2 = −1nC 93 = +1nC The correct ranking cannot be determined. Reset Help most positive

Part A Find the x-component of the electric field at the origin, point O. Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz = Η ΑΣΦ ? N/C Submit Part B Now, assume that charge q2 is negative; q2 = -6 nC, as shown in (Figure 2). What is the x-component of the net electric field at the origin, point O? Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz= Η ΑΣΦ ? N/C

Answer 2

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 3

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 4

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 5

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 6

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 7

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

Answer 8

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 13

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 14

(b)

To determine

The time averaged Poynting vector.

Answer 15

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 20

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 21

(c)

To determine

The instantaneous and time average Poynting vector.

Answer 22

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.