FUNDAMENTALS OF PHYSICS
FUNDAMENTALS OF PHYSICS
12th Edition
ISBN: 9781119798606
Author: Halliday
Publisher: WILEY
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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
Check Mark
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.

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Chapter 23 Solutions

FUNDAMENTALS OF PHYSICS

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