$${\text{For,}}\,r = \frac{R}{2}$$
Using Gauss's law, we have
$$\eqalign{ & \oint {\vec E \cdot d\vec A = \frac{{{q_{{\text{in}}}}}}{{{ \in _0}}}\,\,{\text{or}}\,\,E \times 4\pi {r^2} = \int\limits_0^{\frac{R}{2}} {\frac{{\rho 4\pi {r^2}dr}}{{{ \in _0}}}} } \cr & {\text{or}}\,\,E = \frac{{C{r^3}}}{{5{ \in _0}}} = \frac{{C{R^3}}}{{40{ \in _0}}}. \cr} $$

Both the charges are identical and placed symmetrically about $$ABCD.$$  The flux crossing $$ABCD$$  due to each charge is $$\frac{1}{6}\left[ {\frac{q}{{{ \in _0}}}} \right]$$  but in opposite directions. Therefore the resultant is zero.

Charge on the element opposite to the gap is

$$\eqalign{ & dq = \frac{Q}{{2\pi r}}\left( {0.002\pi } \right) \cr & = \frac{1}{{2\pi \left( {0.5} \right)}} \times \frac{{2\pi }}{{1000}} = 2 \times {10^{ - 3}}C \cr & E = \frac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 3}}}}{{{{\left( {0.5} \right)}^2}}} = 7.2 \times {10^7}N{C^{ - 1}} \cr} $$

The flux is zero according to Gauss’ Law because it is a open surface which enclosed a charge $$q.$$

$$\eqalign{ & \tau = \left( {2.57 \times {{10}^{ - 17}}\,Cm} \right)\left( {3.0 \times {{10}^4}\frac{N}{C}} \right)\left( {\frac{1}{2}} \right) \cr & = 3.855 \times {10^{ - 13}}\,Nm. \cr} $$

$$\eqalign{ & \frac{{ - K2q}}{{{{\left( {x - L} \right)}^2}}} + \frac{{K8q}}{{{x^2}}} = 0 \Rightarrow \frac{1}{{{{\left( {x - L} \right)}^2}}} = \frac{4}{{{x^2}}} \cr & {\text{or,}}\,\frac{1}{{x - L}} = \frac{2}{x} \Rightarrow x = 2x - 2L\,{\text{or}}\,x = 2L \cr} $$

The pattern of field lines shown in option (C) is correct because
(A) a current carrying toroid produces magnetic field lines of such pattern
(B) a changing magnetic field with respect to time in a region perpendicular to the paper produces induced electric field lines of such pattern.