Two metallic parallel plates which have opposite charges, are when placed at a small distance then this arrangement is called capacitor.

Plates are connected with battery of 1000 V.

So, potential difference between the two plates is 1000 Volt.

Given that the distance between the plates d is 2 cm= .02 m

Electric field E= V/d

E= 1000/.02  =50,000 N/C

and Force F= E×q           

Here q is charge on a proton= 1.6×10^-19 C

So,force on proton is = 50,000×1.6×10^-19

= 8×10^-15  N