Since charge Q is outside the hemispherical surface, the net flux passing through the curved surface of hemispherical surface and flat surface is zero.

Therefore,

$$\phi$$_curved + $$\phi$$_flat = 0 ....... (1)

Hence, option (B) is incorrect.

Now,

$${\phi _{flat}} = \int {\overrightarrow E .\,d\overrightarrow A = \int {EdA\cos \theta } } $$

and $$E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{(\sqrt {{R^2} + {r^2}} )}^2}}} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {({R^2} + {r^2})}}$$

Also,

$$\cos \theta = {R \over {\sqrt {{R^2} + {r^2}} }}$$

and

$$A = 2\pi r \Rightarrow dA = 2\pi rdr$$

Therefore,

$${\phi _{flat}} = \int\limits_0^R {{1 \over {4\pi {\varepsilon _0}}}{Q \over {{R^2} + {r^2}}}2\pi rdr{R \over {\sqrt {{R^2} + {r^2}} }}} $$

$${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{rdr} \over {{{({R^2} + {r^2})}^{3/2}}}}} $$

Substituting R² + r² = t, we get

$$2rdr = dt$$

$$ \Rightarrow {\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{dt} \over 2}{1 \over {{t^{3/2}}}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{t^{ - 1/2}}} \over { - 1/2}}} \right]_0^R} $$

Substituting $$t = {R^2} + {r^2}$$, we get

$${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{{({R^2} + {r^2})}^{ - 1/2}}} \over { - 1/2}}} \right]_0^R$$

$$ = {{QR} \over {2{\varepsilon _0}}}\left[ {{{ - 1} \over {\sqrt {{R^2} + {r^2}} }}} \right]_0^R = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {{R^2} + {R^2}} }} + {1 \over {\sqrt {{R^2}} }}} \right)$$

$$ = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {2{R^2}} }} + {1 \over {\sqrt {{R^2}} }}} \right) = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 R}} + {1 \over R}} \right)$$

$$ = {{QR} \over {2{\varepsilon _0}}}{1 \over R}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right) = {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right)$$

Using Eq. (1), we get

$${\phi _{curved}} = - {\phi _{flat}} = - {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right) = - {Q \over {2{\varepsilon _0}}}\left( {1 - {1 \over {\sqrt 2 }}} \right)$$

Hence, option (A) is correct.

The potential at any point on the circumference of the flat surface is $${1 \over {4\pi {\varepsilon _0}}}{Q \over {\sqrt 2 R}}$$.

Thus, the circumference of flat surface is equipotential.