Answer:
The y-component of the electric force on this charge is
![F_y = -1.144* 10^(-6)\ N.](https://img.qammunity.org/2020/formulas/physics/college/n9lafn9859c762z0xlwqoax0hzlkettje7.png)
Step-by-step explanation:
Given:
- Electric field in the region,
![\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.](https://img.qammunity.org/2020/formulas/physics/college/bd8qg18l0k4q4866u9wl28ps6uq9qu4je7.png)
- Charge placed into the region,
![q = 10.4\ nC = 10.4* 10^(-9)\ C.](https://img.qammunity.org/2020/formulas/physics/college/batw883cxa3avkzphexhd2353r5qe6zym9.png)
where,
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,
![\vec E = (\vec F)/(q)\\\therefore \vec F = q\vec E\\=(10.4* 10^(-9))* (148.0\ \hat i-110.0\ \hat j)\\=(1.539* 10^(-6)\ \hat i-1.144* 10^(-6)\ \hat j)\ N.](https://img.qammunity.org/2020/formulas/physics/college/q6l4dnl2lc2gnctqg8pu1keyerg0uuetpc.png)
Thus, the y-component of the electric force on this charge is
![F_y = -1.144* 10^(-6)\ N.](https://img.qammunity.org/2020/formulas/physics/college/n9lafn9859c762z0xlwqoax0hzlkettje7.png)