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:

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

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:
\[
T > mg
\]

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.

To maximize the force between two charges \( q_1 \) and \( q_2 \) obtained by dividing a charge \( Q \) into two parts, we can use Coulomb's law:

\[
F = k \frac{q_1 q_2}{r^2}
\]

where \( r \) is the distance between them.

Steps:

1. Set up the variables:
Let \( q_1 = x \) and \( q_2 = Q - x \).

2. Express the force:
Substitute \( q_1 \) and \( q_2 \) into the formula:
\[
F = k \frac{x (Q - x)}{r^2}
\]

3. Maximize \( F \):
To find the maximum force, take the derivative of \( F \) with respect to \( x \) and set it to zero:
\[
\frac{dF}{dx} = k \frac{Q - 2x}{r^2} = 0
\]

Solving \( Q - 2x = 0 \) gives \( x = \frac{Q}{2} \).

4. Conclusion:
The force is maximum when each part is \( \frac{Q}{2} \).

To find the time period of oscillation of the third charge \(-q\) when displaced perpendicular to the line joining the two charges of magnitude \(+Q\), we can follow these steps:

1. Restoring Force: When the charge \(-q\) is displaced by a small distance \(y\) perpendicular to the line joining the two charges, the net force on it due to the two fixed charges \(+Q\) has a restoring nature and acts toward the equilibrium position.

2. Approximation: For small \(y\), the restoring force \(F\) is proportional to \(y\), making it a simple harmonic motion (SHM) problem.

3. Electric Field: The electric field at the midpoint due to each charge \(+Q\) is \(E = \frac{Q}{4\pi\epsilon_0 (a^2 + y^2)}\). Using \(y \ll a\), we approximate the force on \(-q\) by expanding for small \(y\), resulting in a force \(F = -k y\), where \(k = \frac{2Qq}{4\pi \epsilon_0 a^3}\).

4. Angular Frequency: For SHM, the angular frequency \(\omega\) is \(\sqrt{\frac{k}{m}} = \sqrt{\frac{2Qq}{4\pi \epsilon_0 a^3 m}}\).

5. Time Period: The time period \(T\) is given by:
\[
T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2ma^3 \pi \epsilon_0}{Qq}}
\]