Explain the purpose of wooden stands in electrostatics experiments and calculate charge distribution after conducting spheres touch.

The spheres were placed on wooden stands because wood is an electrical insulator. This prevents charge from flowing from the spheres to the ground, ensuring the spheres retain their charges.

When identical conducting spheres touch, the total charge is shared equally between them.
Initial charge on R, $Q_R = +8 \, \mu C$.
Initial charge on S, $Q_S = -4 \, \mu C$.
Total charge $Q_{total} = Q_R + Q_S = +8 \, \mu C + (-4 \, \mu C) = +4 \, \mu C$.

Since the spheres are identical, the total charge is divided equally.
Charge on each sphere after contact, $Q' = \frac{Q_{total}}{2} = \frac{+4 \, \mu C}{2} = +2 \, \mu C$.
The net charge on each sphere is $\boxed{\text{}+2 \, \mu C\text{}}$ after contact and separation.

After contact, both spheres R and S have a positive charge of $+2 \, \mu C$. Electric field lines originate from positive charges and point outwards. Since both spheres are positively charged, the field lines will repel each other in the region between the spheres, creating a region of weaker field or a neutral point directly between them. The field lines will curve away from each other.

Sphere T has a charge of $+1 \, \mu C$. Sphere R has a charge of $+2 \, \mu C$. Sphere S has a charge of $+2 \, \mu C$.

• Force on T due to R ($F_{RT}$): Since both R and T are positive, the force is repulsive. $F_{RT}$ acts on T, pushing it away from R (to the right).
• Force on T due to S ($F_{ST}$): Since both S and T are positive, the force is repulsive. $F_{ST}$ acts on T, pushing it away from S (to the left).

$Q_R = +2 \, \mu C = 2 \times 10^{-6} \, C$
$Q_S = +2 \, \mu C = 2 \times 10^{-6} \, C$
$Q_T = +1 \, \mu C = 1 \times 10^{-6} \, C$
$r_{RT} = 10 \, cm = 0.1 \, m$
$r_{TS} = 20 \, cm = 0.2 \, m$
Coulomb's constant $k = 9 \times 10^9 \, N \cdot m^2/C^2$.

Using Coulomb's Law, $F = k \frac{|q_1 q_2|}{r^2}$:

Let's take the direction to the right as positive.

The net electrostatic force experienced by T is $\boxed{\text{1.35 N to the right}}$.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge placed at that point. It is a vector quantity.
Mathematically, $E = \frac{F}{q}$.

Using the formula $E = k \frac{|Q|}{r^2}$:

Let's take the direction to the right as positive.

The magnitude of the net electric field at the location of T is \boxed{\text{1.35 \times 10^6 N/C}}.