Solution:

Given:
Mass of the particle, m
Charge on the particle, q1
Fixed charge, -q2
Radius of the circular path, r

To find:
Period of revolution, T
Speed of the particle, v

Formula used:

Coulomb's law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = k(q1*q2)/r^2

where,
k = Coulomb's constant = 9 x 10^9 Nm^2/C^2
q1 and q2 are the charges
r is the distance between the charges

Centripetal force is given by:

Fc = mv^2/r

where,
m is the mass of the particle
v is the speed of the particle
r is the radius of the circular path

Equating the two forces, we get:

k(q1*q2)/r^2 = mv^2/r

Simplifying the above equation, we get:

v = sqrt(k*q1*q2/mr)

Period of revolution is given by:

T = 2πr/v

Substituting the value of v in the above equation, we get:

T = 2πmr/sqrt(k*q1*q2)

Calculation:

Given,
m = 2 x 10^-3 kg
q1 = 3 x 10^-6 C
q2 = -2 x 10^-6 C
r = 0.1 m

Using the formula v = sqrt(k*q1*q2/mr), we get:

v = sqrt(9 x 10^9 * 3 x 10^-6 * -2 x 10^-6 / 2 x 10^-3 * 0.1) = 3 x 10^3 m/s

Using the formula T = 2πmr/sqrt(k*q1*q2), we get:

T = 2π * 2 x 10^-3 * 0.1 / sqrt(9 x 10^9 * 3 x 10^-6 * -2 x 10^-6) = 0.14 s

Therefore, the period of revolution is 0.14 s and the speed of the particle is 3 x 10^3 m/s.