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:

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

To find the relation between \( q_1 \) and \( q_2 \) for \( q_3 \) to be in equilibrium, let's analyze the forces on \( q_3 \).

1. Forces on \( q_3 \):
- The force due to \( q_1 \) on \( q_3 \) is:
\[
F_{13} = \frac{k |q_1| |q_3|}{(2x)^2} = \frac{k |q_1| |q_3|}{4x^2}
\]
(directed towards \( q_1 \) if \( q_1 \) and \( q_3 \) have opposite signs).

- The force due to \( q_2 \) on \( q_3 \) is:
\[
F_{23} = \frac{k |q_2| |q_3|}{x^2}
\]
(directed towards \( q_2 \) if \( q_2 \) and \( q_3 \) have opposite signs).

2. Equilibrium Condition for \( q_3 \):
For \( q_3 \) to be in equilibrium, these forces must be equal in magnitude:
\[
F_{13} = F_{23}
\]
Substituting the values:
\[
\frac{k |q_1| |q_3|}{4x^2} = \frac{k |q_2| |q_3|}{x^2}
\]
Cancelling \( k \), \( |q_3| \), and \( x^2 \) from both sides, we get:
\[
\frac{|q_1|}{4} = |q_2|
\]
Therefore:
\[
q_1 = -4q_2
\]

Answer:
The relation between \( q_1 \) and \( q_2 \) is:
\[
q_1 = -4q_2
\]

Let's solve this briefly.

1. Charges and Positions:
- Charge on \( A = 8 \times 10^{-6} \, \text{C} \).
- Charge on \( B = -2 \times 10^{-6} \, \text{C} \).
- Distance between \( A \) and \( B = 20 \, \text{cm} \).

2. Equilibrium Condition for Third Charge \( Q \):
- For the third charge \( Q \) to experience zero net force, it must be positioned where the attractive force from \( A \) and the repulsive force from \( B \) on \( Q \) are equal in magnitude.
- Since \( |A| > |B| \), \( Q \) must be placed on the right side of \( B \) (further from \( A \)).

3. Distance Calculation:
- Let the distance of \( Q \) from \( B \) be \( x \).
- The distance of \( Q \) from \( A \) is then \( 20 + x \).

Using Coulomb's law, we set the magnitudes of the forces equal:
\[
\frac{k \cdot (8 \times 10^{-6}) \cdot Q}{(20 + x)^2} = \frac{k \cdot (2 \times 10^{-6}) \cdot Q}{x^2}
\]
Simplify by cancelling \( k \) and \( Q \):
\[
\frac{8}{(20 + x)^2} = \frac{2}{x^2}
\]
Cross-multiplying gives:
\[
8x^2 = 2(20 + x)^2
\]
Simplifying, we find \( x = 20 \, \text{cm} \).

Answer:
The third charge should be placed 20 cm to the right of \( B \) for zero net force.

The initial force \( F \) between two charges \( q_1 \) and \( q_2 \) is given by:

\[
F = k \frac{q_1 q_2}{r^2} = 100 \, \text{N}
\]

After changing the charges:
- \( q_1 \) increases by 10%, so \( q_1' = 1.1 q_1 \).
- \( q_2 \) decreases by 10%, so \( q_2' = 0.9 q_2 \).

The new force \( F' \) is:

\[
F' = k \frac{q_1' q_2'}{r^2} = k \frac{(1.1 q_1)(0.9 q_2)}{r^2} = (1.1 \times 0.9) F
\]

Calculating \( 1.1 \times 0.9 = 0.99 \), so:

\[
F' = 0.99 \times 100 = 99 \, \text{N}
\]

Solution Outline:

1. Forces Acting on \( q_1 \):
- Gravitational Force: \( F_g = mg \), acting vertically downward.
- Electrostatic Force: \( F_e = \frac{k |q_1 q_2|}{r^2} \), acting horizontally toward \( q_2 \).

2. Resultant Tension \( T \)
Since \( q_1 \) is in equilibrium, the tension \( T \) in the thread must balance both the gravitational and electrostatic forces. The tension will act diagonally, with components to counteract both the vertical \( mg \) and horizontal \( F_e \) forces.

3. Calculating \( T \):
Using vector addition, we find \( T \) as:
\[
T = \sqrt{(mg)^2 + \left( \frac{k |q_1 q_2|}{r^2} \right)^2}
\]

Since \( \frac{k |q_1 q_2|}{r^2} \) is a positive quantity, \( T \) is indeed greater than \( mg \).

Conclusion:
The correct answer is:
\[
T > mg
\]

To make the oil drop move upward with the same terminal velocity it had while moving downward, the electric force upward must balance the gravitational force downward.

Solution Steps:

1. Given Data:
- Mass of the drop, \( m = 1.6 \times 10^{-12} \, \text{g} = 1.6 \times 10^{-15} \, \text{kg} \)
- Charge on the drop, \( q = 6 \times e = 6 \times 1.6 \times 10^{-19} \, \text{C} = 9.6 \times 10^{-19} \, \text{C} \)
- Gravitational force, \( F_{\text{gravity}} = mg \), where \( g = 9.8 \, \text{m/s}^2 \)

2. Gravitational Force:
\[
F_{\text{gravity}} = (1.6 \times 10^{-15}) \times 9.8 = 1.568 \times 10^{-14} \, \text{N}
\]

3. Electric Field Required:
- To counteract this gravitational force, the electric force \( F_{\text{electric}} = qE \) must equal \( F_{\text{gravity}} \).
\[
qE = F_{\text{gravity}}
\]
\[
E = \frac{F_{\text{gravity}}}{q} = \frac{1.568 \times 10^{-14}}{9.6 \times 10^{-19}} = 3.3 \times 10^4 \, \text{N/C}
\]

Conclusion:
The required electric field magnitude is \( 3.3 \times 10^4 \, \text{N/C} \).

From Newton's Third Law of Motion forces acting on both the charges are equal.

\[ tan \theta = \frac{F}{mg} \]

So angle made by them with vertical will also be equal. But the value of θ will increase because force is increasing due to increase in charge.

To find \(\sum \overrightarrow{F_i}\), the vector sum of forces acting on each charge in the system, we can apply Newton's third law:

1. Newton's Third Law:
- For any pair of charges, the force exerted by one charge on another is equal in magnitude and opposite in direction to the force exerted by the second charge on the first.

2. Net Force on System:
- Since each pair of charges in the system exerts equal and opposite forces on each other, all internal forces cancel out in pairs.

3. Resultant Force:
- As a result, the net force on the entire system of charges is zero.

Conclusion:

The total force \(\sum \overrightarrow{F_i} = 0\).

To analyze the motion of the positive charge \( +q \) released from point \( (2a, 0) \), let's consider the forces acting on it due to the two fixed negative charges \( -q \) at points \( (0, a) \) and \( (0, -a) \).

Solution Steps:

1. Symmetry of Forces:
- Each negative charge \( -q \) exerts an attractive force on \( +q \) along the line joining them.
- Due to symmetry, the vertical components of these forces (along the y-axis) will cancel out, while the horizontal components (along the x-axis) will add up, pulling \( +q \) toward the y-axis.

2. Resultant Force Direction:
- The resultant force always points toward the y-axis but varies with distance from it.
- The force does not follow a linear restoring force law (i.e., it’s not proportional to displacement), which is required for simple harmonic motion (SHM).

3. Motion of \( +q \):
- The charge \( +q \) will oscillate back and forth across the y-axis due to the attractive forces from the two fixed charges.
- However, since the force is not proportional to displacement, the motion is **not SHM**.

Conclusion:
The positive charge \( +q \) will oscillate but will not execute SHM.