A surface has the area vector A → = (2 i ^ + 3 j ^ ) m 2 . What is the flux of a uniform electric field through the area if the field is (a) E → = 4 i ^ N/C and (b) E → = 4 k ^ N/C?

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 23, Problem 1Q

A surface has the area vector A → = (2 i ^ + 3 j ^ ) m². What is the flux of a uniform electric field through the area if the field is (a) E → = 4 i ^ N/C and (b) E → = 4 k ^ N/C?

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Checkpoint 4 The figure shows four orientations of an electric di- pole in an external electric field. Rank the orienta- tions according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the di- pole, greatest first. (1) (2) E (4)

What is integrated science. What is fractional distillation What is simple distillation

19:39 · C Chegg 1 69% ✓ The compound beam is fixed at Ę and supported by rollers at A and B. There are pins at C and D. Take F=1700 lb. (Figure 1) Figure 800 lb ||-5- F 600 lb بتا D E C BO 10 ft 5 ft 4 ft-—— 6 ft — 5 ft- Solved Part A The compound beam is fixed at E and... Hình ảnh có thể có bản quyền. Tìm hiểu thêm Problem A-12 % Chia sẻ kip 800 lb Truy cập ) D Lưu of C 600 lb |-sa+ 10ft 5ft 4ft6ft D E 5 ft- Trying Cheaa Những kết quả này có hữu ích không? There are pins at C and D To F-1200 Egue!) Chegg Solved The compound b... Có Không ☑ ||| Chegg 10 וח

Answer 2

Textbook Question

Chapter 23, Problem 1Q

A surface has the area vector A → = (2 i ^ + 3 j ^ ) m². What is the flux of a uniform electric field through the area if the field is (a) E → = 4 i ^ N/C and (b) E → = 4 k ^ N/C?

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 23, Problem 1Q

A surface has the area vector A → = (2 i ^ + 3 j ^ ) m². What is the flux of a uniform electric field through the area if the field is (a) E → = 4 i ^ N/C and (b) E → = 4 k ^ N/C?

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 23, Problem 1Q

A surface has the area vector A → = (2 i ^ + 3 j ^ ) m². What is the flux of a uniform electric field through the area if the field is (a) E → = 4 i ^ N/C and (b) E → = 4 k ^ N/C?

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Answer 8

Expert Solution & Answer

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ

Where E . → - Electric field

A=Area of given surface

θ- Angle θ between E→ and normal to the surface

3) Given

Area vector, dA→= 2i^+ 3j^m2

Electric field vector in case a, E→= 4i^ N/C

Electric field vector in case b, E→= 4k N/C

4) Calculations

The electric flux through the given surface, ϕ= E . → dA→

a. If E→= 4i^ N/C and dA→= 2i^+ 3j^m2

Plugging the values of E → and dA→ , we obtain

=4i^ N/C . 2i^+ 3j^m2

= 8 N. m2/C

ϕ=8 N. m2/C

b. If E→= 4k N/C and dA→= 2i^+ 3j^m2

As dA does not have k components, the dot product of E→ and dA→ becomes zero.

Therefore,

ϕ=0 N. m2/C

Conclusion:

Electric flux through the given surface is found using the values of dA→ and E→.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To find:

What is the flux of a uniform electric field through the area if the field

a. E→= 4i^ N/C

b. E→= 4k N/C

Answer 12

Answer to Problem 1Q

Solution:

the flux of a uniform electric field through the area if the field

E→= 4i^ N/C is 8 N. m2/C

E→= 4k N/C is 0 N. m2/C

Answer 13

Explanation of Solution

1) Concept

The electric flux through a surface is the amount of electric field that penetrates the surface. Electric flux through a given element with area vector A→ is the dot product of electric field vector and the area vector. Where, the area vector is a vector perpendicular to the surface and has magnitude equal to A.

2) Formulae

Electric flux, ϕ= E . → A→ or E A Cos θ