Yes, the charge q placed at the center of a ring with uniform charge distribution will experience a force.

Explanation:
When a charge q is placed at the center of the ring, it creates an electric field at that point due to the charge distribution on the ring. This electric field exerts a force on the charge q, causing it to experience a net force.

Force Calculation:
The force experienced by the charge q can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

In this case, the charge q at the center of the ring experiences a force due to the electric field created by the charge distribution on the ring. The magnitude of the force is given by:

F = k * |q| * |Q| / r^2

Where:
- F is the force experienced by the charge q
- k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2)
- |q| is the magnitude of the charge q
- |Q| is the total charge on the ring
- r is the radius of the ring

The direction of the force depends on the sign of the charge q and the charge distribution on the ring. If the charge q is positive, it will experience a force in the direction of the electric field lines. If the charge q is negative, it will experience a force in the opposite direction of the electric field lines.

Conclusion:
In conclusion, a charge q placed at the center of a ring with uniform charge distribution will experience a force due to the electric field created by the charge distribution on the ring. The magnitude and direction of the force can be calculated using Coulomb's Law.