Equipotential surfaces associated with an electric field which is increasing in magnitude along the $x$-direction are

Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct?

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$[A]$ The electric flux passing through the curved surface of the hemisphere is $-\frac{\mathrm{Q}}{2 \varepsilon_0}\left(1-\frac{1}{\sqrt{2}}\right)$

$[B]$ Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_0}$

$[C]$ The component of the electric field normal to the flat surface is constant over the surface

$[D]$ The circumference of the flat surface is an equipotential

Assertion : Two equipotential surfaces cannot cut each other.

Reason : Two equipotential surfaces are parallel to each other.

Figure shows a set of equipotential surfaces. The magnitude and direction of electric field that exists in the region is .........