Solution:

The electric flux is given by:

\[
\Phi = \overrightarrow{E} \cdot \overrightarrow{S}
\]

Here, \(\overrightarrow{E} = 2\hat{i} + 4\hat{j} + 3\hat{k}\) and \(\overrightarrow{S} = 10\hat{j}\).

\[
\Phi = (2\hat{i} + 4\hat{j} + 3\hat{k}) \cdot (10\hat{j}) = 0 + 4 \cdot 10 + 0 = 40 \, \text{units}.
\]

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Solution:

The electric flux through a spherical surface depends only on the total enclosed charge, as per Gauss's law:

\[
\Phi = \frac{Q_{\text{enc}}}{\epsilon_0}
\]

Since the total charge enclosed remains the same regardless of the sphere's radius, the flux through the concentric sphere of radius 20 cm will also be 25 V-m.

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Solution:

Angle between area vector and Electric field is 90º. So Flux is zero.

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Solution:

To calculate the electric flux through the lateral surface of the vessel, we apply Gauss's law and consider the symmetry of the system.

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Step-by-Step Solution:

1. Total Flux from the Charge \(q\):
The total flux emitted by the point charge \(q\) in all directions is:
\[
\Phi_{\text{total}} = \frac{q}{\varepsilon_0}.
\]

2. Flux Through the Circular Mouth:
Using the geometry of the vessel, the charge \(q\) is placed at a depth \(h = \sqrt{3}R\). The flux through the circular mouth of radius \(R\) can be calculated as:
\[
\Phi_{\text{mouth}} = \frac{q}{2\varepsilon_0}.
\]

3. Flux Through the Lateral Surface
By symmetry, the flux through the lateral surface is the remaining flux from the total flux after subtracting the flux through the mouth:
\[
\Phi_{\text{lateral}} = \Phi_{\text{total}} - \Phi_{\text{mouth}}.
\]

4. Simplify:
Substitute the values:
\[
\Phi_{\text{lateral}} = \frac{q}{\varepsilon_0} - \frac{q}{2\varepsilon_0}.
\]
\[
\Phi_{\text{lateral}} = \frac{q}{2\varepsilon_0} \left(1 + \frac{\sqrt{3}}{2}\right).
\]

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Final Answer:
The electric flux through the lateral surface of the vessel is:
\[
{\frac{q}{2\varepsilon_0} \left( 1 + \frac{\sqrt{3}}{2} \right)}.
\]

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Solution:

We are tasked with finding the charge enclosed within the cube using Gauss's Law:

\[
\Phi_{\text{net}} = \frac{q_{\text{enc}}}{\varepsilon_0}.
\]

Given:
- Net flux through the cube: \(\Phi_{\text{net}} = 1.05 \, \text{Nm}^2\text{C}^{-1}\),
- \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}\),
- \(\alpha = 800 \, \text{N}\text{C}^{-1}\text{m}^{-1/2}\),
- Cube side length: \(a = 0.1 \, \text{m}\).

Steps:

1. Use Gauss's Law:
The enclosed charge \(q_{\text{enc}}\) is related to the flux by:
\[
q_{\text{enc}} = \varepsilon_0 \Phi_{\text{net}}.
\]

2. Substitute the values:
\[
q_{\text{enc}} = (8.85 \times 10^{-12}) \cdot (1.05).
\]

3. Calculate:
\[
q_{\text{enc}} = 9.27 \times 10^{-12} \, \text{C}.
\]

Final Answer:
The enclosed charge is:
\[
{9.27 \times 10^{-12} \, \text{C}}.
\]

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Solution:

Step-by-Step Explanation:

1. Flux through the Entire Pyramid:
The total charge enclosed by the pyramid is \(+Q\), and the total flux through the entire closed surface of the pyramid is given by Gauss's law:
\[
\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}.
\]

2. Flux Distribution:
The pyramid has a square base and four triangular faces. However, **the charge \(+Q\) is located at the center of the base, not at the geometric center of the pyramid.** This means the flux through the base is not zero and contributes to the total flux.

3. Flux Through the Base:
Due to symmetry, half of the total flux passes through the square base:
\[
\Phi_{\text{base}} = \frac{\Phi_{\text{total}}}{2} = \frac{Q}{2\varepsilon_0}.
\]

4. Flux Through the Four Triangular Faces:
The remaining half of the total flux passes through the four triangular faces combined:
\[
\Phi_{\text{triangular (total)}} = \frac{\Phi_{\text{total}}}{2} = \frac{Q}{2\varepsilon_0}.
\]

5. Flux Through One Triangular Face:
Since the four triangular faces are identical, the flux is equally distributed among them:
\[
\Phi_{\text{face}} = \frac{\Phi_{\text{triangular (total)}}}{4} = \frac{\frac{Q}{2\varepsilon_0}}{4} = \frac{Q}{8\varepsilon_0}.
\]

Final Answer:
The flux through one triangular face is:
\[
{\frac{Q}{8\varepsilon_0}}.
\]

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Solution:

To hold equal charges \(q\) in equilibrium at the corners of a square, we need to place a charge \(Q\) at the center of the square such that the net force on each corner charge due to other charges is balanced.

1. Force due to corner charges:
Each charge \(q\) at the corners experiences repulsive forces due to the other three corner charges. The net force from these three charges is:
\[
F_{\text{corners}} = q \cdot \frac{1}{4\pi\varepsilon_0} \left( \frac{2}{a^2} + \frac{\sqrt{2}}{a^2} \right) = \frac{q^2}{4\pi\varepsilon_0 a^2} \left( 2 + \sqrt{2} \right),
\]
where \(a\) is the side of the square.

2. Force due to center charge \(Q\):
The attractive force due to the central charge \(Q\) on each corner charge is:
\[
F_{\text{center}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r^2},
\]
where \(r = \frac{a}{\sqrt{2}}\) is the distance from the center to a corner. Substituting \(r\):
\[
F_{\text{center}} = \frac{qQ}{4\pi\varepsilon_0 \cdot \frac{a^2}{2}} = \frac{2qQ}{4\pi\varepsilon_0 a^2}.
\]

3. Equilibrium condition:
For equilibrium, \(F_{\text{corners}} = F_{\text{center}}\):
\[
\frac{q^2}{4\pi\varepsilon_0 a^2} \left( 2 + \sqrt{2} \right) = \frac{2qQ}{4\pi\varepsilon_0 a^2}.
\]

4. Solve for \(Q\):
\[
Q = \frac{q}{2} \left( 2 + \sqrt{2} \right).
\]
Since \(Q\) is opposite in sign to \(q\), the final charge is:
\[
Q = -\frac{q}{4} \left( 1 + 2\sqrt{2} \right).
\]

Thus, the required charge at the center is:

\[
Q = -\frac{q}{4} \left( 1 + 2\sqrt{2} \right).
\]

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Solution:

The total flux due to a charge \(Q\) is given by Gauss's law:

\[
\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}.
\]

Since the charge \(Q\) is placed at the mouth of the conical flask, the flux through the flask corresponds to half the total flux (because the charge is symmetrically distributed above and below the mouth):

\[
\Phi_{\text{flask}} = \frac{1}{2} \cdot \Phi_{\text{total}} = \frac{1}{2} \cdot \frac{Q}{\varepsilon_0}.
\]

Thus, the flux through the conical flask is:

\[
\Phi = \frac{Q}{2\varepsilon_0}.
\]

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Solution:

The electric flux is given by:

\[
\Phi = \vec{E} \cdot \vec{A} = E A \cos\theta,
\]

where:
- \(E = 2 \times 10^3 \, \text{V/m}\),
- \(A = \text{area of the coil} = 10 \, \text{cm} \times 20 \, \text{cm} = 0.1 \, \text{m} \times 0.2 \, \text{m} = 0.02 \, \text{m}^2\),
- \(\theta = 0^\circ\) (field is perpendicular to the coil, as the coil is in the \(xy\)-plane and the field is along \(\hat{k}\)).

Substitute values:
\[
\Phi = (2 \times 10^3) \cdot (0.02) \cdot \cos(0^\circ),
\]
\[
\Phi = 40 \, \text{V·m}.
\]

Thus, the electric flux is:

\[
40 \, \text{V·m}
\]

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Solution:

The electric flux \(\Phi\) through the Gaussian surface can be found using Gauss's law:

\[
\Phi = \frac{q_{\text{enc}}}{\epsilon_0}
\]

Here, \(q_{\text{enc}}\) is the charge enclosed by the spherical Gaussian surface.

Since the infinite sheet has a uniform surface charge density \(\sigma\), the charge enclosed is the product of \(\sigma\) and the area of the sheet that lies inside the sphere. The area of the sheet within the sphere is the area of the circle formed by the intersection, which has a radius \(r\) determined by the geometry of the sphere and the plane.

1. The radius of the intersection circle is \(r = \sqrt{R^2 - x^2}\).
2. The area of this circle is \(A = \pi r^2 = \pi (R^2 - x^2)\).

Thus, the enclosed charge is:

\[
q_{\text{enc}} = \sigma A = \sigma \pi (R^2 - x^2).
\]

Substituting this into Gauss's law:

\[
\Phi = \frac{\sigma \pi (R^2 - x^2)}{\epsilon_0}.
\]

So, the electric flux through the Gaussian surface is:

\[
\Phi = \frac{\sigma \pi (R^2 - x^2)}{\epsilon_0}.
\]

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