A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 1*24.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by E → = (3.00 + 2.00x 2 )i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?

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Answer 1

Textbook Question

Chapter 24, Problem 24.63CP

A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 1*24.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by $E →$ = (3.00 + 2.00x²)i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?

Chapter 24, Problem 24.63CP, A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

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Answer 2

Textbook Question

Chapter 24, Problem 24.63CP

A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 1*24.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by $E →$ = (3.00 + 2.00x²)i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?

Chapter 24, Problem 24.63CP, A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

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Answer 3

Textbook Question

Chapter 24, Problem 24.63CP

A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 1*24.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by $E →$ = (3.00 + 2.00x²)i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?

Chapter 24, Problem 24.63CP, A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

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Answer 4

Textbook Question

Chapter 24, Problem 24.63CP

A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 1*24.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by $E →$ = (3.00 + 2.00x²)i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?

Chapter 24, Problem 24.63CP, A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

Answer 8

(a)

Expert Solution

To determine

The net electric flux leaving the closed surface.

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The net electric flux leaving the closed surface.

Answer 13

Answer to Problem 24.63CP

The net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Answer 14

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The position of the closed surface is,

Physics For Scientists And Engineers, Technology Update, Loose-leaf Version, Chapter 24, Problem 24.63CP

Figure (1)

From the figure (1) the electric field is perpendicular to the normal of area ABEF, area CDHG, area ABCD and the area EFGH. Therefore,

The formula to calculate electric flux is,

$ϕ=EAcosθ$ (1)

Here,

$E$ is the magnitude of the electric field crossing the area.

$A1$ is the area of the closed surface.

$θ$ is the angle between the normal to surface area of rectangular plane and the electric field.

The electric field makes an angle $90°$ with the normal of area ABCD, area EFGH, area ABEF and area CDHG.

Substitute $90°$ for $θ$ in the equation (1).

$ϕ=EAcos90°=0$

Thus, the flux out of the surface ABEF, CDHG, ABCD and EFGH is zero for the surfaces.

From figure (1), the area of the surface ACGE is,

$AACGE=ab$

From figure (1), the area of the surface BFDH is,

$ABFDH=ab$

The electric flux in the $x$ -direction is,

$E=(3.00+2.00x2) N/C$ (2)

From figure (1), substitute $a$ for $x$ in the above equation to calculate the electric field at surface of area ACGE.

$E=(3.00+2.00(a)2) N/C$

From Figure (1), substitute $a+c$ for $x$ in the equation (2) to calculate the electric field at surface of area BFDH.

$E=(3.00+2.00(a+c)2) N/C$

The net flux leaving closed surface,

$ϕnet=EBFDHABFDH−EACGEAACGE$

Here,

$EBFDH$ is the electric field that leaves the surface BFDH.

$EACGE$ is the electric field that enters the surface ACGE.

Substitute $(3.00+2.00(a+c)2) N/C$ for $EBFDH$ and $(3.00+2.00(a)2) N/C$ for $EACGE$ , $ab$ for $ABFDH$ and $ab$ for $AACGE$ in the above equation.

$ϕnet=((3.00+2.00(a+c)2) N/C)ab−((3.00+2.00(a)2) N/C)ab=((3.00+2.00(a+c)2) N/C−(3.00+2.00(a)2) N/C)ab=(2.00((a+c)2−(a)2) N/C)ab=(2.00(a+c−a)(a+c+a) N/C)ab$

Simplify the above equation.

$ϕnet=(2.00(c)(2a+c) N/C)ab=(2.00)(2a+c)abc N/C$

Substitute $0.400 m$ for $a$ , $0.400 m$ for $b$ and $0.600 m$ for $c$ in the above equation.

$ϕnet=(2.00 N/C)(2(0.400 m)+(0.600 m))(0.400 m)(0.400 m)(0.600 m)=0.2688 Nm2/C≈0.269 N⋅m2/C$

Conclusion:

Therefore, the net electric flux leaving the closed surface is $0.269 N⋅m2/C$ .

Answer 15

(b)

Expert Solution

To determine

The net charge enclosed by the surface.

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The net charge enclosed by the surface.

Answer 20

Answer to Problem 24.63CP

The net charge enclosed by the surface is $2.38×10−12 C$ .

Answer 21

Explanation of Solution

Given info: The dimensions of the closed surface are $a=b=0.400 m$ and $c=0.600 m$ and the electric field is given by $E→=(3.00+2.00x2)i^ N/C$ .

The flux enclosed by a closed surface is given by,

$ϕ=qencε0$

Here,

$qenc$ is the charge enclosed inside the closed surface.

$ϕ$ is the net flux through the closed surface.

$ε0$ is the permittivity due to space.

The permittivity of free space is $8.85×10−12 C2/Nm2$ .

Substitute $0.269 Nm2/C$ for $ϕ$ and $8.85×10−12 C2/Nm2$ for $ε0$ in the above equation.

$0.269 Nm2/C=qenc8.85×10−12 C2/Nm2$

Rearrange the above equation for $qenc$ .

$qenc=(0.269 Nm2/C)(8.85×10−12 C2/Nm2)=2.38×10−12 C$

Conclusion:

Therefore, the net charge enclosed by the surface is $2.38×10−12 C$ .

Answer 22

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Answer to Problem 24.63CP

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Answer to Problem 24.63CP

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