Certain equipotential surfaces are depicted in the figure....

In situation (a), the equipotential lines are parallel, indicating a uniform electric field.

The perpendicular distance $\Delta r$ between the $10\text{ V}$ and $20\text{ V}$ lines is $\Delta r = (20\text{ cm} - 10\text{ cm}) \sin 30^\circ = 10 \times 0.5 = 5\text{ cm} = 0.05\text{ m}$.

Magnitude of electric field $E = \dfrac{\Delta V}{\Delta r} = \dfrac{20 - 10}{0.05} = 200\text{ V m}^{-1}$.

The electric field is perpendicular to the equipotential lines and points towards decreasing potential (negative x-direction). Thus, the angle with the positive x-axis is $30^\circ + 90^\circ = 120^\circ$.

In situation (b), the equipotential surfaces are concentric circles. Observing the given values with $r$ in meters:

At $r = 0.1\text{ m}$, $V = 60\text{ V} \Rightarrow V \cdot r = 6$

At $r = 0.2\text{ m}$, $V = 30\text{ V} \Rightarrow V \cdot r = 6$

At $r = 0.3\text{ m}$, $V = 20\text{ V} \Rightarrow V \cdot r = 6$

Thus, the potential follows the relation $V = \dfrac{6}{r}$.

The magnitude of the electric field is $E = -\dfrac{dV}{dr} = -\dfrac{d}{dr}\left(\dfrac{6}{r}\right) = \dfrac{6}{r^2}$.

Since the potential decreases as the radial distance increases, the electric field points radially outward.

Answer: (a) $200 \text{ V m}^{-1}$ making an angle of $120^\circ$ with the $x$-axis; (b) radially outward, decreasing as $E = \dfrac{6 \text{ V m}}{r^2}$