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

(B) Equation $A$ is $\oint E \cdot dA = \frac{Q}{\varepsilon_0}$,which is the Gauss law for electrostatics. It states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.
Equation $B$ is $\oint B \cdot dA = 0$,which is the Gauss law for magnetism. It implies that magnetic monopoles do not exist,meaning the net magnetic flux through any closed surface is always zero.

(C) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $Q_{\text{enclosed}}$ inside the surface divided by the permittivity of free space $\epsilon_0$.
Mathematically,$\phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
Since the charge $Q$ is placed at the centre of the cube,the entire charge is enclosed by the cube.
Therefore,the total flux through the six surfaces of the cube is $\phi = \frac{Q}{\epsilon_0}$.

(A, B, D) $1$. Gauss's Law is derived from Coulomb's Law and the superposition principle. Therefore,it inherently contains both of these fundamental principles of electrostatics.
$2$. The area element $d\vec{s}$ (or $d\vec{A}$) is defined as a vector pointing outward,normal to the surface. Since it is defined as the cross product of two displacement vectors (e.g.,$d\vec{l}_1 \times d\vec{l}_2$),it behaves as a pseudo-vector (or axial vector) because its direction depends on the orientation of the surface coordinate system,not just the position.

(A, C) According to Gauss's Law,the total electric flux $\phi_{\text{total}}$ through any closed surface in a region with no enclosed charge is zero,i.e.,$\phi_{\text{total}} = 0$.
For the hemisphere,$\phi_{\text{total}} = \phi_{\text{curved}} + \phi_{\text{flat}} = 0$.
The flux through the flat circular surface is $\phi_{\text{flat}} = -E \cdot A = -E(\pi R^2)$ (since the field lines enter the surface).
Therefore,$\phi_{\text{curved}} = -\phi_{\text{flat}} = E \pi R^2$.
Thus,statement $(A)$ is correct and $(C)$ is correct.
Regarding the work done,for a uniform electric field,the potential difference $\Delta V = -\vec{E} \cdot \Delta \vec{r}$.
Since points $A$ and $B$ lie on the same equipotential plane perpendicular to the electric field $\vec{E}$,the potential difference between $A$ and $B$ is $\Delta V = 0$.
Therefore,the work done $W = q \Delta V = 0$,which is independent of the path and the radius $R$.

(C) $1$. The charge $+Q$ at the center $O$ contributes a total flux of $\frac{Q}{\varepsilon_{0}}$ through the entire cube. By symmetry,the flux through each of the $6$ faces is $\frac{Q}{6\varepsilon_{0}}$.
$2$. The charge $+Q$ at point $P$ is located at a distance $a/2$ from the center of the face $ABCD$. This charge creates an electric field. The flux due to charge $+Q$ at $P$ through the face $ABCD$ is $\frac{Q}{2\varepsilon_{0}}$ (since the charge is at a distance $a/2$ from the center of the face,it subtends a solid angle of $2\pi$ steradians at the face,or by considering an identical cube placed adjacent to the first one).
$3$. The net flux through face $ABCD$ due to both charges is $\phi_{ABCD} = \phi_{O, ABCD} + \phi_{P, ABCD} = \frac{Q}{6\varepsilon_{0}} - \frac{Q}{2\varepsilon_{0}} = -\frac{Q}{3\varepsilon_{0}}$ (the flux from $P$ enters the cube,hence negative).
$4$. Total flux through the cube due to both charges is $\phi_{total} = \frac{Q_{enclosed}}{\varepsilon_{0}} = \frac{Q}{\varepsilon_{0}}$.
$5$. The flux through the remaining $5$ faces is $\phi_{5} = \phi_{total} - \phi_{ABCD} = \frac{Q}{\varepsilon_{0}} - (-\frac{Q}{3\varepsilon_{0}}) = \frac{4Q}{3\varepsilon_{0}} = \frac{8Q}{6\varepsilon_{0}}$.

(A) According to Gauss's law,the total electric flux $\phi_{total}$ through any closed surface is given by $\phi_{total} = \frac{Q}{\varepsilon_{0}}$,where $Q$ is the net charge enclosed by the surface.
Since the charge $Q$ is placed at the centre of the cube,the electric flux is symmetrically distributed through all $6$ faces of the cube.
Therefore,the electric flux $\phi_{face}$ through any one face of the cube is $\phi_{face} = \frac{\phi_{total}}{6} = \frac{Q}{6 \varepsilon_{0}}$.

(D) Given,electric field intensity $E = (2i + 3j + k) \ NC^{-1}$.
Area vector $S = 10i \ m^2$.
The electric flux $\phi$ through a surface is defined by the dot product of the electric field vector and the area vector:
$\phi = E \cdot S$
Substituting the given values:
$\phi = (2i + 3j + k) \cdot (10i)$
Since $i \cdot i = 1$,$j \cdot i = 0$,and $k \cdot i = 0$:
$\phi = (2 \times 10) + (3 \times 0) + (1 \times 0)$
$\phi = 20 \ Nm^2 C^{-1}$.

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$.
In this problem,the point charge $+q$ is placed at the centre of the cube,which is a closed surface.
Therefore,the total electric flux emerging from the cube is $\phi = \frac{q}{\varepsilon_0}$.

(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is $\phi = \frac{Q_{\text{in}}}{\varepsilon_0}$,where $Q_{\text{in}}$ is the net charge enclosed by the surface.
$1$. The charge $2q$ is placed at vertex $A$. $A$ vertex of a cube is shared by $8$ adjacent cubes. Therefore,the fraction of charge $2q$ enclosed within the cube is $\frac{2q}{8} = \frac{q}{4}$.
$2$. The charge $q$ is placed at the centre of the face $CDEF$. $A$ face of a cube is shared by $2$ adjacent cubes. Therefore,the fraction of charge $q$ enclosed within the cube is $\frac{q}{2}$.
$3$. The total enclosed charge $Q_{\text{in}}$ is the sum of these contributions: $Q_{\text{in}} = \frac{q}{4} + \frac{q}{2} = \frac{q + 2q}{4} = \frac{3q}{4}$.
$4$. Thus,the total electric flux passing through the cube is $\phi = \frac{Q_{\text{in}}}{\varepsilon_0} = \frac{3q}{4\varepsilon_0}$.

(B) According to Gauss's law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_0}$.
When a charge $q$ is placed at a vertex of a cube,the cube occupies only $\frac{1}{8}$ of the total space surrounding the charge because $8$ identical cubes are required to enclose the charge completely at the center of a larger cube.
Therefore,the electric flux passing through the cube is $\Phi = \frac{1}{8} \times \frac{q}{\varepsilon_0} = \frac{q}{8\varepsilon_0}$.

(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$.
For the first sphere with radius $r_1 = 3 \text{ cm}$,only the charge at $x = 2 \text{ cm}$ $(q_1 = 8 \mu \text{C})$ is enclosed within the sphere.
Therefore,the flux through the first sphere is $\phi_1 = \frac{8 \mu \text{C}}{\epsilon_0}$.
For the second sphere with radius $r_2 = 5 \text{ cm}$,both charges at $x = 2 \text{ cm}$ $(8 \mu \text{C})$ and $x = 4 \text{ cm}$ $(-2 \mu \text{C})$ are enclosed within the sphere.
Therefore,the total enclosed charge is $q_{\text{total}} = 8 \mu \text{C} - 2 \mu \text{C} = 6 \mu \text{C}$.
The flux through the second sphere is $\phi_2 = \frac{6 \mu \text{C}}{\epsilon_0}$.
The ratio of the electric flux is $\frac{\phi_1}{\phi_2} = \frac{8 \mu \text{C} / \epsilon_0}{6 \mu \text{C} / \epsilon_0} = \frac{8}{6} = \frac{4}{3}$.