The electric field \( E \) is related to the potential difference \( \Delta V \) by:

\[
\Delta V = -\int_{x_1}^{x_2} E \, dx
\]

Given \( E = \frac{100}{x^2} \), the potential difference between \( x = 10 \) m and \( x = 20 \) m is:

\[
\Delta V = -\int_{10}^{20} \frac{100}{x^2} \, dx
\]

\[
\Delta V = -\left[ \frac{100}{x} \right]_{10}^{20}
\]

\[
\Delta V = -\left( \frac{100}{20} - \frac{100}{10} \right)
\]

\[
\Delta V = -\left( 5 - 10 \right) = 5 \, \text{V}
\]

Thus, the potential difference is \( 5 \, \text{V} \).

Electric potential decreases in the direction of Electric Field. So Field will be maximum at A

Work done in a conservative field is not dependent on path taken by the object so,

\[ dV= - \overrightarrow{E}.\overrightarrow{dr}\]

\[ V= - E\widehat{i}.(-a\widehat{i}-b\widehat{j})\]

so, Potential difference = -E.a

Work Done= -Q. E.a

On interchanging A and B with C and D direction of electric field will get reversed. But potential being scaler quantity remains unchanged.

To determine the electric field \(\vec{E}\) from the equipotential surfaces:

1. Electric Field Magnitude: The electric field is the negative gradient of the potential:
\[
|\vec{E}| = -\frac{\Delta V}{\Delta s}
\]
where \(\Delta V\) is the potential difference and \(\Delta s\) is the perpendicular spacing between equipotential surfaces.

2. Horizontal Component (\(E_x\)):
- The horizontal spacing between 0V and 10V surfaces is \(2 \, \text{units}\).
- \(E_x = -\frac{\Delta V_x}{\Delta x} = -\frac{10}{2} = -5 \, \text{V/unit}\).

3. Vertical Component (\(E_y\)):
- The vertical spacing between equipotential surfaces (tilted at \(37^\circ\)) corresponds to \(\tan 37^\circ = \frac{3}{4}\). Hence, the vertical spacing between 0V and 10V is \(2 \cdot \frac{3}{4} = 1.5 \, \text{units}\).
- \(E_y = -\frac{\Delta V_y}{\Delta y} = -\frac{10}{1.5} = -\frac{20}{3} \, \text{V/unit}\).

4. Resultant Electric Field:
\[
\vec{E} = -E_x \hat{i} - E_y \hat{j} = 5\hat{i} + \frac{20}{3}\hat{j}.
\]

Electric Potential (\( V \)):

1. Potential at the center due to a charge \( q \) at a distance \( r \) is:
\[
V = \frac{kq}{r}.
\]

2. Total potential is the algebraic sum since potential is scalar. For 3 positive charges and 5 negative charges:
\[
V_{\text{total}} = 3 \cdot \frac{kq}{r} - 5 \cdot \frac{kq}{r} = \frac{-2kq}{r}.
\]

---

Electric Field (\( E \)):

1. Symmetry of arrangement: The charges along the diagonals cancel their horizontal components, leaving only vertical components.

2. Field due to charges along diagonals (\( -q, -q, +q, +q \)):
- Each diagonal pair creates a net field of magnitude:
\[
E_{\text{pair}} = \frac{\sqrt{2}kq}{r^2}.
\]
- Two such pairs contribute:
\[
E_{\text{diagonals}} = 2 \cdot \frac{\sqrt{2}kq}{r^2} = \frac{2\sqrt{2}kq}{r^2}.
\]

3. Field due to the remaining vertical pair (\( -q, -q \)):
- Net field:
\[
E_{\text{vertical}} = \frac{2kq}{r^2}.
\]

4. Total electric field:
\[
E_{\text{total}} = E_{\text{diagonals}} + E_{\text{vertical}} = \frac{2kq}{r^2}(1 + \sqrt{2}).
\]

---

Final Answer:
\[
E = \frac{2kq}{r^2}(1 + \sqrt{2}), \quad V = \frac{-2kq}{r}.
\]

To find the ratio \( q/Q \), we need to calculate the total potential energy of the system and set it to zero. The charges are arranged in a straight line as shown:

Charges:
1. \( -q \) at one end.
2. \( +Q \) in the middle.
3. \( -q \) at the other end.

Distances:
- Distance between \( -q \) and \( +Q \): \( x \).
- Distance between \( +Q \) and the other \( -q \): \( x \).
- Total distance between the two \( -q \) charges: \( 2x \).

Potential Energy of the System:
The potential energy \( U \) for each pair of charges is given by:
\[
U = \frac{kq_1q_2}{r},
\]
where \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant.

1. Interaction between \( -q \) and \( +Q \) (on one side):
\[
U_1 = \frac{k(-q)(+Q)}{x} = -\frac{kqQ}{x}.
\]

2. Interaction between \( +Q \) and \( -q \) (on the other side):
\[
U_2 = \frac{k(+Q)(-q)}{x} = -\frac{kqQ}{x}.
\]

3. Interaction between \( -q \) and \( -q \) (across the distance \( 2x \)):
\[
U_3 = \frac{k(-q)(-q)}{2x} = \frac{kq^2}{2x}.
\]

Total Potential Energy:
\[
U_{\text{total}} = U_1 + U_2 + U_3 = -\frac{kqQ}{x} - \frac{kqQ}{x} + \frac{kq^2}{2x}.
\]
Simplify:
\[
U_{\text{total}} = -\frac{2kqQ}{x} + \frac{kq^2}{2x}.
\]

Set \( U_{\text{total}} = 0 \):
\[
-\frac{2kqQ}{x} + \frac{kq^2}{2x} = 0.
\]

Factor out \( \frac{k}{x} \):
\[
-\frac{2qQ}{1} + \frac{q^2}{2} = 0.
\]

Multiply through by 2 to eliminate the fraction:
\[
-4qQ + q^2 = 0.
\]

Factorize:
\[
q(q - 4Q) = 0.
\]

Since \( q \neq 0 \), we have:
\[
q = 4Q.
\]

Ratio:
\[
\frac{q}{Q} = 4.
\]

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

---=====================================

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

----------------------------------------------------------------------------------------

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

To find the magnitude of the third charge that will keep the system in equilibrium, follow these steps:

Setup:
- Two charges \( Q \) are placed at a distance \( d \) apart.
- A third charge \( q \) is placed at the midpoint of the line joining the two charges.

For the system to be in equilibrium, the net force on all the charges must be zero.

Forces on the third charge \( q \) at the midpoint:
- The two charges \( Q \) exert forces on \( q \) from opposite directions.
- These forces will be equal in magnitude but opposite in direction, so we need to balance them by choosing the right value for \( q \).

Using Coulomb's law, the force between \( Q \) and \( q \) (at a distance \( d/2 \)) is given by:

\[
F = k \frac{|Q \cdot q|}{(d/2)^2} = k \frac{4|Q \cdot q|}{d^2}
\]

Forces on one of the charges \( Q \) (let's take the left charge):
- The charge \( q \) at the midpoint exerts a force on \( Q \) in one direction.
- The other charge \( Q \) exerts a repulsive force on this charge from the opposite direction.

For equilibrium, the force between the two \( Q \)'s must balance the force due to \( q \) on \( Q \).

1. Force between the two \( Q \)'s:

\[
F_{QQ} = k \frac{Q^2}{d^2}
\]

2. Force between \( Q \) and \( q \) (distance \( d/2 \)):

\[
F_{Qq} = k \frac{4|Q \cdot q|}{d^2}
\]

Condition for equilibrium:
The force between the two \( Q \)'s must equal the force between \( Q \) and \( q \):

\[
k \frac{Q^2}{d^2} = k \frac{4|Q \cdot q|}{d^2}
\]

Canceling out the common terms:

\[
Q^2 = 4|Q \cdot q|
\]

\[
q = \frac{Q}{4}
\]

Conclusion:
The magnitude of the third charge \( q \) that should be placed at the midpoint for equilibrium is:

\[
q = \frac{- Q}{4}
\]