Answer
See the detailed answer below.
Work Step by Step
From the given graph, we can see that:
- $B=-0.10\;\hat z\;\rm T$
- $E=-10^6\;\hat i\;\rm V/m$
- $v=10^7\;\hat i\;\rm m/s$
So the net force exerted on the charge is given by
$$\sum F=q(\vec E+\vec v\times \vec B)$$
Plug the given;
$$\sum F=(1.6\times 10^{-19})(-10^6\;\hat i+[10^7\;\hat i\times-0.10\;\hat z])$$
$$\sum F=(1.6\times 10^{-19})(-10^6\;\hat i+10^6\;\hat j)$$
$$\sum F= (- 1.6\times 10^{-13}\;\hat i+1.6\times 10^{-13}\;\hat j)\;\rm N$$
The magnitude of this force is given by
$$\left|\sum F\right|=\sqrt{(- 1.6\times 10^{-13})^2+( 1.6\times 10^{-13})^2}$$
$$\left|\sum F\right|=\color{red}{\bf 2.3\times 10^{-13}}\;\rm N$$
It angle relative to positive $y$-directino is given by
$$\tan\theta=\dfrac{F_x}{F_y}$$
$$\theta=\tan^{-1}\left[\dfrac{1.6\times 10^{-13}}{1.6\times 10^{-13}}\right]$$
$$\theta=\color{red}{\bf45}^\circ\tag {CCW from $+y$-direction}$$
