A small ball of mass 20 g carries a charge of +5.0 µC…

Step 1. Equilibrium angle (a). The electric force is $qE = 5.0 \times 10^{-6}\ \text{C} \times 2.94 \times 10^{4}\ \text{N/C} = 0.147\ \text{N}$ and the weight is $mg = 0.196\ \text{N}$. The thread lies along their sum, so $\tan\theta = \dfrac{qE}{mg} = 0.75$ and $\theta = 36.9^\circ$.

Step 2. Tension (b). $T = \sqrt{(mg)^2 + (qE)^2} = 0.245\ \text{N}$. So the tilted gravity is $g_{\text{eff}} = T/m = 12.25\ \text{m/s}^2$.

Step 3. The swing (c). Released at the vertical, the ball starts at rest, $36.9^\circ$ to one side of its equilibrium line. A pendulum swings as far on the other side, so the thread reaches $2\theta = 73.7^\circ$ from the vertical.

Step 4. Greatest speed (d). The ball is fastest on the equilibrium line. Measured along $g_{\text{eff}}$, it has dropped by $L(1 - \cos\theta)$ to get there:

Step 5. Greatest tension (e). On the equilibrium line the thread must supply $mg_{\text{eff}}$ and the centripetal force: $T = 0.245\ \text{N} + \dfrac{mv^2}{L} = 0.245\ \text{N} + 0.098\ \text{N} = 0.343\ \text{N}$.

Step 6. Small oscillations (f). $T_{\text{osc}} = 2\pi\sqrt{\dfrac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\dfrac{0.80\ \text{m}}{12.25\ \text{m/s}^2}} = 1.61\ \text{s}$.