Answer:
The y-component of the electric force on this charge is [tex]F_y = -1.144\times 10^{-6}\ N.[/tex]
Explanation:
Given:
where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
[tex]\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.[/tex]
Thus, the y-component of the electric force on this charge is [tex]F_y = -1.144\times 10^{-6}\ N.[/tex]