Question 1

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge -Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

Accepted Answer

According to Gauss's law, the electric flux through a closed surface depends only on the charge enclosed, not the size of the surface. Since both surfaces enclose the same charge, the flux is the same.

Question 2

Which one of the following cannot be a statement of Gauss's Law for some physical situation?

Accepted Answer

Gauss's law states that the net electric flux through any closed surface is proportional to the net charge enclosed by the surface. Option D does not satisfy this condition as it does not include the enclosed charge.

Question 3

A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge.What is the electric flux through one face of the cube?

Accepted Answer

The electric flux through one face of the cube can be calculated using the formula:
Electric flux = electric field x surface area x cos(theta)
Since the charge is at the center of the cube, the electric field will be uniform and point radially outwards from the center. The magnitude of the electric field can be calculated using Coulomb's law:
E = k*q/r^2
where k is the Coulomb constant (9x10^9 Nm^2/C^2), q is the charge (0.8 nC or 8x10^-10 C), and r is the distance from the charge to any face of the cube (which is half the length of a diagonal of the cube, or 2sqrt(3) m).
So the electric field at any face of the cube is:
E = (9x10^9 Nm^2/C^2)*(8x10^-10 C)/(2sqrt(3) m)^2
E = 2.6x10^6 N/C
The surface area of any face of the cube is (4 m)^2 = 16 m^2.
Since the electric field is radial, the angle theta between the electric field and the surface normal of any face is 0 degrees. Therefore, cos(theta) = 1.
Putting it all together, the electric flux through one face of the cube is:
Electric flux = (2.6x10^6 N/C)*(16 m^2)*(1)
Electric flux = 41.6x10^6 Nm^2/C = 15 Nm^2/C = 15 N/C
Therefore, the best choice is B.

Question 4

The xy plane is "painted" with a uniform surface charge density which is equal to 40 nC/m².Consider a spherical surface with a 4.0-cm radius that has a point in the xy plane as its center.What is the electric flux through that part of the spherical surface for which z > 0?

Accepted Answer

The answer of The xy plane is "painted" with a...

Question 5

A uniform linear charge density of 4.0 nC/m is distributed along the entire x axis.Consider a spherical (radius = 5.0 cm)surface centered on the origin.Determine the electric flux through this surface.

Accepted Answer

The answer of A uniform linear charge density of 4.0...

Question 6

The electric flux through the two adjacent spherical surfaces shown below is known to be the same.  It is also known that there is no charge inside either spherical surface.We can conclude that

Accepted Answer

The answer of The electric flux through the two adjacent...

Question 7

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

Accepted Answer

According to Gauss's law, the electric flux through a closed surface depends only on the charge enclosed, not the size of the surface. Since both surfaces enclose the same charge Q, the flux through both surfaces is the same.

Question 8

A long cylinder (radius = 3.0 cm)is filled with a nonconducting material which carries a uniform charge density of 1.3 $\mu$C/m³.Determine the electric flux through a spherical surface (radius = 2.0 cm)which has a point on the axis of the cylinder as its center.

Accepted Answer

The electric flux through the spherical surface is given by:
ϕ = Qenclosed/ε0
where Qenclosed is the charge enclosed within the spherical surface and ε0 is the permittivity of free space.

Since the cylinder has a uniform charge density, the total charge enclosed by the spherical surface can be calculated as:
Qenclosed = ρV
where ρ is the charge density, and V is the volume enclosed by the spherical surface.

The volume of the cylinder enclosed by the spherical surface can be calculated as:
V = πh(r^2 - h^2/3)
where h is the distance between the center of the sphere and the plane of the ends of the cylinder, r is the radius of the cylinder. In this case, h = 3 cm and r = 3 cm.

Substituting the values, we get:
V = π(3 cm)(3^2 cm^2 - (3 cm)^2/3) = 54π cm^3

The charge enclosed within the spherical surface is:
Qenclosed = (1.3 μC/m^3)(54π cm^3) = 69.66 μC

Substituting back into the expression for electric flux, we get:
ϕ = (69.66 μC)/(8.85 x 10^-12 C^2/N*m^2) = 4.9 x 10^11 N*m^2/C or 4.9 N*m^2/C

Therefore, the best choice is B.

Question 9

A charge of uniform volume density (40 nC/m³)fills a cube with 8.0-cm edges.What is the total electric flux through the surface of this cube?

Accepted Answer

The answer of A charge of uniform volume density (40...

Question 10

You are told that $\oint \overrightarrow { \mathbf { E } } \cdot d \overrightarrow { \mathbf { A } }$ summed over both the surface areas of sphere A and sphere B below totals to $\frac { - Q } { \varepsilon _ { 0 } }$ .You can conclude that

Accepted Answer

The answer of You are told that \(\oint \overrightarrow {...

Question 11

A uniform charge density of 500 nC/m³ is distributed throughout a spherical volume (radius = 16 cm).Consider a cubical (4.0 cm along the edge)surface completely inside the sphere.Determine the electric flux through this surface.

Accepted Answer

The electric flux through the cubical surface is determined by the charge enclosed within it. The charge density is 500 nC/m³, and the volume of the cube is (0.04 m)³. The enclosed charge is then multiplied by the charge density, and the flux is calculated using Gauss's law.

Question 12

A point charge +Q is located on the x axis at x = a,and a second point charge -Q is located on the x axis at x = -a.A Gaussian surface with radius r = 2a is centered at the origin.The flux through this Gaussian surface is

Accepted Answer

By symmetry, the electric field at any point on the Gaussian surface will have the same magnitude and direction for opposite points on the surface. Therefore, the flux over one hemisphere will be equal and opposite to the flux over the other hemisphere, resulting in a total flux of zero.

Question 13

Charge of uniform surface density (4.0 nC/m²)is distributed on a spherical surface (radius = 2.0 cm).What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?

Accepted Answer

The total electric flux through a spherical surface is given by Φ = Q/ε₀ where Q is the total charge enclosed by the surface and ε₀ is the electric constant. In this case, the enclosed charge is the product of the surface charge density (σ) and the surface area of the spherical shell between the two surfaces (4πr²). Therefore, the enclosed charge is Q = σ(4πr²) = (4.0 nC/m²)(4π(0.04 m)²) = 1.024 nC. The electric flux through the larger surface is then Φ = Q/ε₀A = (1.024 nC)/(8.85×10⁻¹² C²/N⋅m²)(4π(0.04 m)²) = 2.28×10⁹ N⋅m²/C = 2.3 N⋅m²/C (or 2.3 V⋅m). Therefore, choice C is the correct answer.

Question 14

Which one of the following is not an expression for electric charge?

Accepted Answer

The expression $\int \mathbf { E } \cdot d \mathbf { A }$ represents electric flux, not electric charge.

Question 15

If we define the gravitational field $\overrightarrow { \mathbf { g } } = \frac { - G M _ { \text {source } } } { r ^ { 2 } } \hat { \mathbf { r } }$ ,where $\hat{r}$ is a unit radial vector,then Gauss's Law for gravity is

Accepted Answer

Gauss's Law for gravity states that the flux of the gravitational field through a closed surface is proportional to the enclosed mass, with a factor of $-4\pi G$.

Question 16

A hemispherical surface (half of a spherical surface)of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere.What is the magnitude of the electric flux through the hemisphere surface?

Accepted Answer

The answer of A hemispherical surface (half of a spherical...

Question 17

Gino says that the analog of Gauss's law for the flow of an incompressible fluid of density$\nu$ at constant velocity $\overrightarrow { \mathbf { v } }$ is $\oint \rho\overrightarrow { \mathbf { v } } \cdot d \mathbf { A } = 0$ for an imaginary surface within the fluid.Lorenzo says that it is true only if the area where the fluid enters the surface and the area where it leaves the surface are both perpendicular to the velocity of the fluid.Which one,if either,is correct?

Accepted Answer

Gino is correct because for an incompressible fluid at constant velocity, the net flow across any closed surface is zero, as the amount of fluid entering equals the amount leaving, regardless of the angle.

Question 18

A point charge is located at the origin.Centered along the x axis is a cylindrical closed surface of radius 10 cm with one end surface located at x = 2 m and the other end surface located at x = 4 m.If the magnitude of the electric flux through the surface at x = 2 m is 4 N .m²/C,what is the magnitude of the electric flux through the surface at x = 4 m?

Accepted Answer

The answer of A point charge is located at the...

Question 19

The electric field in the region of space shown is given by $$\overrightarrow { \mathbf { E } } = ( 8 \tilde { \mathbf { i } } + 2 \hat { \mathbf { j } } )$$ N/C where y is in m.What is the magnitude of the electric flux through the top face of the cube shown?

Accepted Answer

The electric flux through the top face of the cube is given by the product of the electric field and the area of the face.
The area of the top face is (0.4 m) x (0.4 m) = 0.16 m^2.
So, the electric flux through the top face is:
flux = (electric field) x (area of top face)
= (11 N/C) x (0.16 m^2)
= 1.76 N m^2/C
Thus, the correct answer is C, which is 54 N m^2/C in the desired format.

Question 20

An constant electric field, $$\overrightarrow { \mathbf { E } } = ( 3 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } } )$$ N/C,goes through a surface with area $$\overrightarrow { \mathbf { A } } = ( 8 \hat { \mathbf { i } } - 6 \hat { \mathbf { k } } )$$ m².(This surface can also be expressed as an area of 10 m² with the direction of the unit vector ( $0.8 \tilde { \mathbf { i } } - 0.6 \tilde { \mathbf { k } }$ ).What is the magnitude of the electric flux through this area?

Accepted Answer

The electric flux through a surface is given by the dot product of the electric field and the surface area vector. In this case, we have:$$\overrightarrow { \mathbf { \Phi } } = \overrightarrow { \mathbf { E } } \cdot \overrightarrow { \mathbf { A } } = ( 3 	ilde { \mathbf { i } } + 4 \hat { \mathbf { j } } ) \cdot ( 8 \hat { \mathbf { i } } - 6 \hat { \mathbf { k } } ) = 24 	ext{ N m}^2/	ext{C}$$Therefore, the magnitude of the electric flux through the surface is 24 N m2/C.

Quiz 19: Electric Forces and Electric Fields

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge -Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

Which one of the following cannot be a statement of Gauss's Law for some physical situation?

A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge.What is the electric flux through one face of the cube?

The xy plane is "painted" with a uniform surface charge density which is equal to 40 nC/m².Consider a spherical surface with a 4.0-cm radius that has a point in the xy plane as its center.What is the electric flux through that part of the spherical surface for which z > 0?

A uniform linear charge density of 4.0 nC/m is distributed along the entire x axis.Consider a spherical (radius = 5.0 cm) surface centered on the origin.Determine the electric flux through this surface.

The electric flux through the two adjacent spherical surfaces shown below is known to be the same. It is also known that there is no charge inside either spherical surface.We can conclude that

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

A long cylinder (radius = 3.0 cm) is filled with a nonconducting material which carries a uniform charge density of 1.3 $\mu$ C/m³.Determine the electric flux through a spherical surface (radius = 2.0 cm) which has a point on the axis of the cylinder as its center.

A charge of uniform volume density (40 nC/m³) fills a cube with 8.0-cm edges.What is the total electric flux through the surface of this cube?

A uniform charge density of 500 nC/m³ is distributed throughout a spherical volume (radius = 16 cm) .Consider a cubical (4.0 cm along the edge) surface completely inside the sphere.Determine the electric flux through this surface.

A point charge +Q is located on the x axis at x = a,and a second point charge -Q is located on the x axis at x = -a.A Gaussian surface with radius r = 2a is centered at the origin.The flux through this Gaussian surface is

Charge of uniform surface density (4.0 nC/m²) is distributed on a spherical surface (radius = 2.0 cm) .What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?

Which one of the following is not an expression for electric charge?

If we define the gravitational field $\overrightarrow { \mathbf { g } } = \frac { - G M _ { \text {source } } } { r ^ { 2 } } \hat { \mathbf { r } }$ ,where $\hat{r}$ is a unit radial vector,then Gauss's Law for gravity is

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere.What is the magnitude of the electric flux through the hemisphere surface?

Quiz 19: Electric Forces and Electric Fields

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge -Q located at the center of the surfaces.When compared to the electric flux ϕ\phiϕ 1 through the surface of radius R,the electric flux ϕ\phiϕ 2 through the surface of radius 2R is

Which one of the following cannot be a statement of Gauss's Law for some physical situation?

A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge.What is the electric flux through one face of the cube?

The xy plane is "painted" with a uniform surface charge density which is equal to 40 nC/m2.Consider a spherical surface with a 4.0-cm radius that has a point in the xy plane as its center.What is the electric flux through that part of the spherical surface for which z > 0?

A uniform linear charge density of 4.0 nC/m is distributed along the entire x axis.Consider a spherical (radius = 5.0 cm) surface centered on the origin.Determine the electric flux through this surface.

The electric flux through the two adjacent spherical surfaces shown below is known to be the same. It is also known that there is no charge inside either spherical surface.We can conclude that

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the surfaces.When compared to the electric flux ϕ\phiϕ 1 through the surface of radius R,the electric flux ϕ\phiϕ 2 through the surface of radius 2R is

A long cylinder (radius = 3.0 cm) is filled with a nonconducting material which carries a uniform charge density of 1.3 μ\muμ C/m3.Determine the electric flux through a spherical surface (radius = 2.0 cm) which has a point on the axis of the cylinder as its center.

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges.What is the total electric flux through the surface of this cube?

You are told that ∮E→⋅dA→\oint \overrightarrow { \mathbf { E } } \cdot d \overrightarrow { \mathbf { A } }∮E⋅dA summed over both the surface areas of sphere A and sphere B below totals to −Qε0\frac { - Q } { \varepsilon _ { 0 } }ε0​−Q​ .You can conclude that

A uniform charge density of 500 nC/m3 is distributed throughout a spherical volume (radius = 16 cm) .Consider a cubical (4.0 cm along the edge) surface completely inside the sphere.Determine the electric flux through this surface.

A point charge +Q is located on the x axis at x = a,and a second point charge -Q is located on the x axis at x = -a.A Gaussian surface with radius r = 2a is centered at the origin.The flux through this Gaussian surface is

Charge of uniform surface density (4.0 nC/m2) is distributed on a spherical surface (radius = 2.0 cm) .What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?

Which one of the following is not an expression for electric charge?

If we define the gravitational field g→=−GMsource r2r^\overrightarrow { \mathbf { g } } = \frac { - G M _ { \text {source } } } { r ^ { 2 } } \hat { \mathbf { r } }g​=r2−GMsource ​​r^ ,where r^\hat{r}r^ is a unit radial vector,then Gauss's Law for gravity is

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere.What is the magnitude of the electric flux through the hemisphere surface?

The electric field in the region of space shown is given by E→=(8i~+2j^) \overrightarrow { \mathbf { E } } = ( 8 \tilde { \mathbf { i } } + 2 \hat { \mathbf { j } } ) E=(8i~+2j^​) N/C where y is in m.What is the magnitude of the electric flux through the top face of the cube shown?

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge -Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

The xy plane is "painted" with a uniform surface charge density which is equal to 40 nC/m².Consider a spherical surface with a 4.0-cm radius that has a point in the xy plane as its center.What is the electric flux through that part of the spherical surface for which z > 0?

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the surfaces.When compared to the electric flux $\phi$ ₁ through the surface of radius R,the electric flux $\phi$ ₂ through the surface of radius 2R is

A long cylinder (radius = 3.0 cm) is filled with a nonconducting material which carries a uniform charge density of 1.3 $\mu$ C/m³.Determine the electric flux through a spherical surface (radius = 2.0 cm) which has a point on the axis of the cylinder as its center.

A charge of uniform volume density (40 nC/m³) fills a cube with 8.0-cm edges.What is the total electric flux through the surface of this cube?

A uniform charge density of 500 nC/m³ is distributed throughout a spherical volume (radius = 16 cm) .Consider a cubical (4.0 cm along the edge) surface completely inside the sphere.Determine the electric flux through this surface.

Charge of uniform surface density (4.0 nC/m²) is distributed on a spherical surface (radius = 2.0 cm) .What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?

If we define the gravitational field $\overrightarrow { \mathbf { g } } = \frac { - G M _ { \text {source } } } { r ^ { 2 } } \hat { \mathbf { r } }$ ,where $\hat{r}$ is a unit radial vector,then Gauss's Law for gravity is