EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100460300
Author: SERWAY
Publisher: YUZU
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Chapter 24, Problem 24.42P

In a certain region of space, the electric field is E = 6.00 × 103 x2 i ^ , where E is in newtons per coulomb and x is in meters. Electric charges in this region are at rest and remain at rest. (a) Find the volume density of electric charge at x = 0.300 m. Suggestion: Apply Gauss’s law to a box between x = 0.300 m and x = 0.300 m + dx. (b) Could this region of space be inside a conductor?

(a)

Expert Solution
Check Mark
To determine

The volume density of electric charge at x=0.300m .

Answer to Problem 24.42P

The volume density of electric charge at x=0.300m is 3.19×108C/m2 .

Explanation of Solution

Given info: The electric field vector in the region of space is 6.00×103x2i^ .

Consider a Gaussian box of thickness dx in the given region of space.

Formula to calculate the electric flux at x is equal 3.00m is,

ϕ1=E1Acosθ1

Here,

E1 is the electric field in the region of space at x=0.300m .

A is the normal surface area to the electric field of the Gaussian box.

θ1 is the angle between the electric field and normal surface area of the box.

Write the expression for the electric field at x=0.300m .

E1=6.00×103x2i^

Substitute 0.300m for x in above equation.

E1=6.00×103(0.300m)2i^N/C=540i^N/C

Thus, electric field in the region of space at x=0.300m is 540i^N/C .

The electric field enter into the box at x is equal 3.00m and the normal vector of the area of box is opposite to the direction of the electric field. So, the angle between the electric field and the normal area vector becomes 180° .

Substitute 180° for θ1 and 540N/C for E1 in above equation.

ϕ1=(540N/C)Acos180°=(540N/C)A

Formula to calculate the electric flux at x is equal 3.00m+dx is,

ϕ2=E2Acosθ2

Here,

E2 is the electric field in the region of space at x=0.300m+dx .

θ2 is the angle between the electric field and normal surface area of the box at x=3.00m+dx .

Write the expression for the electric field at x=0.300m+dx .

E2=(6.00×103)(0.300m+dx)2i^N/C=(6.00×103)(0.09m2+dx2+0.6mdx)i^N/C

The term dx2 is very small so it can be neglected in above expression.

E2=(6.00×103)(0.09m2+0.6mdx)i^N/C

The electric field enter into the box at x is equal 3.00m+dx and the normal vector of the area of box is in the same the direction of the electric field. So, the angle between the electric field and the normal area vector becomes 0° .

Substitute 0° for θ2 and (6.00×103)(0.09m2+0.6mdx)N/C for E2 in above equation.

ϕ2=((6.00×103)(0.09m2+0.6mdx)N/C)Acos0°=((6.00×103)(0.09m2+0.6mdx)N/C)A

Write the expression for the net electric flux through the box.

ϕnet=ϕ1+ϕ2

Substitute (540N/C)A for ϕ1 and ((6.00×103)(0.09m2+0.6mdx)N/C)A for ϕ2 in above equation.

ϕnet=(540N/C)A+((6.00×103)(0.09m2+0.6mdx)N/C)A=(3.6×103m)(dx)AN/C (1)

Write an alternate expression for the net electric flux from Gauss law.

ϕnet=qε0 (2)

Here,

q is the average charge enclosed inside the Gaussian box.

ε0 is the permittivity of free space.

Formula to calculate the volume of the Gaussian box is,

V=Adx

Here,

h is the height of the rectangular Gaussian surface.

Formula to calculate the average charge enclosed inside the rectangular Gaussian surface is,

q=ρV

Here,

ρ is the average volume density.

V is the volume of the rectangular Gaussian surface.

Replace V by Adx in above equation.

q=ρAdx

Substitute ρAdx for q in equation (2).

ϕ=ρAdxε0 (3)

Equate the equation (2) and (1) for same value of electric flux.

ρAdxε0=(3.6×103m)(dx)AN/Cρ=(3.6×103m)ε0N/C

Substitute 8.85×1012C2/Nm2 for ε0 .

ρ=(3.6×103N/Cm)×8.85×1012C2/Nm2=3.186×108C/m23.19×108C/m2

Conclusion:

Therefore, the volume density of electric charge at x=0.300m is 3.19×108C/m2 .

(b)

Expert Solution
Check Mark
To determine

Whether the given region of space can be inside a conductor.

Answer to Problem 24.42P

No, the given region of space could not be inside a conductor.

Explanation of Solution

Given info: The electric field vector in the region of space is 6.00×103x2i^ .

According to the principle of Electromagnetism, in electrostatics free charges in a good conductor reside only on the surface. So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is uniform, the electric fields cancel each other. No matter, what is the shape of the conductor as long as there is field inside it, electrons always rearrange themselves to make the net field zero.

But, the volume density of electric charge at x=0.300m is 3.19×108C/m2 , which is a non-zero value. So, the given region of space at x=0.300m could not be inside a conductor.

Conclusion:

Therefore, the given region of space could not be inside a conductor.

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Chapter 24 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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