To find the \( x \)-component of the force on charge \(-q_1\) in the given setup, let's break down the forces due to \( +q_2 \) and \( -q_3 \) on \( -q_1 \).

1. Force due to \( +q_2 \):
The force \( F_{12} \) between \( -q_1 \) and \( +q_2 \) acts along the \( x \)-axis (horizontally) and has a magnitude:
\[
F_{12} = \frac{k |q_1| |q_2|}{b^2}
\]
Since this force acts along the positive \( x \)-direction, the \( x \)-component of this force is simply:
\[
F_{12x} = \frac{k |q_1| q_2}{b^2}
\]

2. Force due to \( -q_3 \):
The force \( F_{13} \) between \( -q_1 \) and \( -q_3 \) acts along the line joining them, which makes an angle \( \theta \) with the \( y \)-axis. The magnitude of this force is:
\[
F_{13} = \frac{k |q_1| |q_3|}{a^2}
\]
To find the \( x \)-component of \( F_{13} \), we use the angle \( \theta \) with the \( y \)-axis. The \( x \)-component of \( F_{13} \) is:
\[
F_{13x} = \frac{k |q_1| |q_3|}{a^2} \sin \theta
\]
(positive, as it points in the \( +x \)-direction due to the symmetry).

3. Total \( x \)-component of the force on \( -q_1 \):
Adding the \( x \)-components of both forces, we get:
\[
F_x = F_{12x} + F_{13x} = \frac{k |q_1| q_2}{b^2} + \frac{k |q_1| |q_3|}{a^2} \sin \theta
\]

Final Answer
After dividing by \( k |q_1| \) to find the proportional relation, the \( x \)-component of the force on \( -q_1 \) is proportional to:
\[
\frac{q_2}{b^2} + \frac{q_3}{a^2} \sin \theta
\]

Given Data
- Two spherical conductors \( B \) and \( C \) have equal radii and equal charges.
- They repel each other with a force \( F \) when kept at a certain distance apart.

Step-by-Step Solution
1. Initial Charge on B and C:
Let's assume the initial charge on both \( B \) and \( C \) is \( q \).
Since they are at a certain distance \( r \) apart, the initial force of repulsion between \( B \) and \( C \) is given by Coulomb's law:
\[
F = \frac{k q^2}{r^2}
\]

2. Introducing the Third Conductor (A):
- A third conductor \( A \) with the same radius as \( B \) and \( C \) is initially uncharged.
- \( A \) is first brought in contact with \( B \), allowing charges to redistribute.

3. Charge Redistribution (First Contact with B):
- When \( A \) (initially uncharged) is brought into contact with \( B \) (which has charge \( q \)), the charge will distribute equally between \( A \) and \( B \) because they have the same radius.
- After contact, the charge on each (both \( A \) and \( B \)) will be:
\[
\frac{q}{2}
\]

4. Charge Redistribution (Then Contact with C):
- Next, \( A \) (which now has charge \( \frac{q}{2} \)) is brought in contact with \( C \) (which has charge \( q \)).
- The charge will again distribute equally between \( A \) and \( C \) because they have the same radius.
- After contact, the charge on each (both \( A \) and \( C \)) will be:
\[
\frac{q}{2} + \frac{q}{2} = \frac{3q}{4}
\]
- So, now \( C \) has a charge of \( \frac{3q}{4} \), and \( A \) also has \( \frac{3q}{4} \).

5. Final Charges on B and C:
- After removing \( A \), the charges on \( B \) and \( C \) are as follows:
- \( B \) has \( \frac{q}{2} \).
- \( C \) has \( \frac{3q}{4} \).

6. New Force of Repulsion between \( B \) and \( C \):
- The new force \( F' \) between \( B \) and \( C \), separated by the same distance \( r \), is given by:
\[
F' = \frac{k \left(\frac{q}{2}\right) \left(\frac{3q}{4}\right)}{r^2}
\]
- Simplifying, we get:
\[
F' = \frac{k \cdot q^2 \cdot 3}{8r^2} = \frac{3}{8} \cdot \frac{k q^2}{r^2}
\]
- Since \( F = \frac{k q^2}{r^2} \), we can write:
\[
F' = \frac{3}{8} F
\]

Final Answer
The new force of repulsion between \( B \) and \( C \) is:
\[
\frac{3F}{8}
\]

Initial Setup:
1. Charges and Forces:
- Two charges \( +q \) and \( +3q \) are separated by distance \( r \).
- The force between them initially is:
\[
F = \frac{k \cdot q \cdot 3q}{r^2} = \frac{3kq^2}{r^2}
\]

2. Accelerations:
- Let the masses of the charges be \( m_1 \) and \( m_2 \).
- Given that the initial accelerations are \( a \) and \( 2a \) for charges \( +q \) and \( +3q \) respectively, we have:
\[
F = m_1 a \quad \text{and} \quad F = m_2 \cdot 2a
\]

Step 1: Determine Mass Ratio
From the equations \( m_1 a = m_2 \cdot 2a \), we get:
\[
m_2 = \frac{m_1}{2}
\]

Step 2: New Charges After Redistribution
When the total charge \( q + 3q = 4q \) is equally distributed, each charge will be:
\[
\frac{4q}{2} = 2q
\]

Step 3: New Force Between Charges
With the new charges \( +2q \) and \( +2q \) at the same distance \( r \), the new force \( F' \) is:
\[
F' = \frac{k \cdot 2q \cdot 2q}{r^2} = \frac{4kq^2}{r^2}
\]

Step 4: New Accelerations
Using \( F' = m_1 a_1' \) and \( F' = m_2 a_2' \), we get:
1. For the first charge (\( m_1 \)):
\[
a_1' = \frac{F'}{m_1} = \frac{4kq^2 / r^2}{m_1} = \frac{4a}{3}
\]

2. For the second charge (\( m_2 = \frac{m_1}{2} \)):
\[
a_2' = \frac{F'}{m_2} = \frac{4kq^2 / r^2}{m_1 / 2} = \frac{8a}{3}
\]

Answer
The new accelerations are:
\[
\frac{4a}{3} \quad \text{and} \quad \frac{8a}{3}
\]

When the spheres have charges \( +Q \) and \( -Q \), they attract each other with force \( \overrightarrow{F} \).

If we replace \( -Q \) with \( +Q \), both spheres will have \( +Q \) charge, causing a repulsive force instead of attraction.

Since \( r > 2R \), there will be induced charges on the surfaces due to electrostatic induction, reducing the effective repulsive force compared to the attractive force \( \overrightarrow{F} \) in the initial configuration.

Conclusion:
The new force will be in the opposite direction to \( \overrightarrow{F} \) (repulsive) but with a magnitude less than \( F \) due to the effect of induced charges.

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:

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

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.