The concept required to perform this exercise is given by the coulomb law.
The force expressed according to this law is given by
[tex]F= \frac{kqQ}{r^2}[/tex]
Where,
[tex]k = 8.99 * 10^9 N m^2 / C^2.[/tex]
q = charges of the objects
r= distance/radius
Our values are previously given, so
[tex]q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59[/tex]
Replacing,
[tex]F=\frac{kqQ}{r^2}[/tex]
[tex]F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}[/tex]
[tex]F= 4.8423N[/tex]
The force acting on the block are given by,
[tex]F-mgsin\theta = ma[/tex]
[tex]a = \frac{F-mgsin\theta}{m}[/tex]
[tex]a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}[/tex][tex]a = 10.31m/s^2[/tex]
Therefore the box is accelerated upward.
