We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

The values of the velocities are,


We know that the acceleration is equivalent to the change of the speed in a certain time therefore



Now applying the Newton's second law we have,



Therefore the approximate magnitude is 8516.36N
Answer:
Explanation:
Given
side of square shape 
Electric flux 
Permittivity of free space 
Flux is given by

where E=electric field strength
A=area
=Angle between Electric field and area vector



and Electric field by a uniformly charged sheet is given by

where
=charge density


More force needs to be applied