The ball is subject to gravitational force $F_g = mg$ (downwards) and electric force $F_e = qE$ (horizontal). For the ball to move in a straight line at 45°, the net force must be collinear with the velocity. This requires the magnitudes of the vertical and horizontal forces to be equal.

The electric field points from high potential (200 V) to low potential (-200 V), i.e., to the left.

$$E = \frac{\Delta V}{d} = \frac{200 \text{ V}}{0.01 \text{ m}} = 20000 \text{ V/m}$$

[Q1] The net force must be directed downwards and to the left (anti-parallel to $v_0$) for deceleration to occur. Thus, the electric force must be to the left. Since $\vec{F}_e = q\vec{E}$ and $\vec{E}$ is to the left, the charge $q$ must be positive. For the net force to be at 45°, we must have $F_e = F_g$.

$$qE = mg \implies q = \frac{mg}{E}$$ $$q = \frac{(0.01 \text{ kg})(10 \text{ m/s}^2)}{20000 \text{ V/m}} = 5 \times 10^{-6} \text{ C}$$

[Q2] The net force is $F_{net} = \sqrt{F_g^2 + F_e^2} = \sqrt{(mg)^2 + (mg)^2} = mg\sqrt{2}$. This force causes a constant deceleration $a = F_{net}/m = g\sqrt{2}$. Using the kinematic equation $v_f^2 = v_i^2 + 2as$, the ball stops when $v_f=0$. The distance $s$ is:

$$0 = v_0^2 - 2as \implies s = \frac{v_0^2}{2a} = \frac{v_0^2}{2g\sqrt{2}}$$ $$s = \frac{(1 \text{ m/s})^2}{2(10 \text{ m/s}^2)\sqrt{2}} = \frac{1}{20\sqrt{2}} \text{ m} = \frac{\sqrt{2}}{40} \text{ m}$$