A force of $$10$$ N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

Between the two plates of a parallel plate capacitor, the electric field is due to both plates. Each plate produces a uniform electric field of $$\frac{\sigma}{2\epsilon_0}$$.

Between the plates, the fields from both plates add up:

$$E_{between} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$

The force on the charged particle between the plates is:

$$F = qE_{between} = \frac{q\sigma}{\epsilon_0} = 10 \text{ N}$$

When one plate is removed, only one plate remains. The electric field due to a single infinite plate is:

$$E_{single} = \frac{\sigma}{2\epsilon_0}$$

The force on the charged particle now becomes:

$$F' = qE_{single} = \frac{q\sigma}{2\epsilon_0} = \frac{F}{2} = \frac{10}{2} = 5 \text{ N}$$

The correct answer is Option A.