The charges make an electric dipole. $$A$$ and $$B$$ points lie on the equatorial plane of the dipole.
Therefore, potential at $$A$$ = potential at $$B = 0$$
$$W = q\left( {{V_A} - {V_B}} \right) = q \times 0 = 0$$

There are 2 forces acting on pendulum (i) its weight

(i) Weight, $$mg = 30.7 \times {10^{ - 6}} \times 9.8$$
$$ = 3 \times {10^{ - 4}}\,N$$    vertically downward, and
(ii) Force, $$F = qE$$
$$\eqalign{ & = 2 \times {10^{ - 8}} \times 20000 \cr & = 4 \times {10^{ - 4}}N \cr} $$
In the horizontal direction.
So, net tension, $$T = \sqrt {{{\left( {mg} \right)}^2} + {F^2}} $$
$$\eqalign{ & = \sqrt {{{\left( {3 \times {{10}^{ - 4}}} \right)}^2} + {{\left( {4 \times {{10}^{ - 4}}} \right)}^2}} \cr & = 5 \times {10^{ - 4}}\,N \cr} $$

Flux does not depend on the size and shape of the close surface, and so, it remains same.

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface = $${\phi _2} - {\phi _1}$$
∴ the charge enclosed in the surface by Gauss’s law is $$q = { \in _0}\left( {{\phi _2} - {\phi _1}} \right)$$

$$\oint {\vec E.d\vec A = 0} ,$$   represents charge inside close surface is zero. Electric field as any point on the surface may be zero.

Electric field everywhere inside the metallic portion of shell is zero.
Hence options (A) and (D) are incorrect.
Electric field lines are always normal to a surface. Hence option (B) is incorrect. Only option (C) represents the correct answer.

Given, $$E = Ar\,......\left( {\text{i}} \right)$$

Here, $$r = a \Rightarrow E = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{q}{{{a^2}}}$$
From Eq. (i),
$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{q}{{{a^2}}} = Aa \Rightarrow q = 4\pi {\varepsilon _0}A{a^3}$$

The electric field inside a thin spherical shell of radius $$R$$ has charge $$Q$$ spread uniformly over its surface is Zero.

Outside the shell the electric field is $$E = k\frac{Q}{{{r^2}}}.$$   These characteristics are represented by graph (A).

Electric field
$$E = - \frac{{d\phi }}{{dr}} = - 2ar\,......\left( {\text{i}} \right)$$
By Gauss's theorem
$$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii),
$$\eqalign{ & q = - 8\pi {\varepsilon _0}a{r^3} \cr & \Rightarrow dq = - 24\pi {\varepsilon _0}a{r^2}dr \cr} $$
Charge density, $$\rho = \frac{{dq}}{{4\pi {r^2}dr}} = - 6{\varepsilon _0}a$$

Potential difference between any two points in an electric field is given by,
$$\eqalign{ & dV = - \overrightarrow E .\overrightarrow {dx} \cr & \int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}dx} } \cr & {V_A} - {V_O} = - \left[ {10{x^3}} \right]_0^2 = - 80J/C \cr} $$