Electric field between two charged parallel plates

For a charged parallel plate capacitor, each plate produces an electric field of magnitude $E_1 = \frac{\sigma}{2\varepsilon_0}$ where $\sigma$ is the surface charge density.

Net field when both plates are present

Inside the capacitor, the fields due to the two plates are in the same direction, so they add: $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$

The force on a charged particle of charge $q$ is $F = qE$

Given: $F = 10\,\text{N}$

When one plate is removed

Then only one charged plate remains, so the electric field becomes $E' = \frac{\sigma}{2\varepsilon_0} = \frac{E}{2}$

Hence the new force is also half: $F' = qE' = q\left(\frac{E}{2}\right) = \frac{F}{2} = \frac{10}{2} = 5\,\text{N}$

Correct option

$\boxed{5\,\text{N}}$

So the correct option is A.