$$\eqalign{ & F = qE = q\left( {A - Bx} \right) \cr & ma = q\left( {A - Bx} \right) \Rightarrow a = \frac{q}{m}\left( {A - Bx} \right)\,......\left( {\text{i}} \right) \cr & \frac{{vdv}}{{dx}} = \frac{q}{m}\left( {A - Bx} \right);vdv = \frac{q}{m}\left( {A - Bx} \right)dx \cr & \int\limits_0^0 {vdv} = \frac{q}{m}\int\limits_0^x {\left( {A - Bx} \right)dx} ;Ax - \frac{{B{x^2}}}{2} = 0 \cr & x = 0,x = \frac{{2A}}{B}\,......\left( {{\text{ii}}} \right) \cr} $$
From eqs. (i) and (ii)
$$\frac{q}{m}\left( {A = Bx} \right) = \frac{q}{m}\left( {A - B \times \frac{{2A}}{B}} \right) = \frac{{ - qA}}{m}.$$

An electric dipole consists of a pair of equal and opposite point charges separated by a very small distance. Atoms or molecules of ammonia, water, alcohol, carbon dioxide, $$HCl$$  etc are some of the examples of electric dipoles, because in these cases, the centres of positive and negative charge distributions ate separated by some small distance.

Here, torque, $$\tau = pE\sin \theta $$

Potential energy of the dipole,
$$\eqalign{ & U = \int {\tau d\theta } = \int_{\frac{\pi }{2}}^0 {pE\sin \theta d\theta = - pE\left[ {\cos \theta } \right]_{\frac{\pi }{2}}^0} \cr & = - pE\cos \theta \cr} $$

Figure shows the electric fields due to the sheets 1, 2 and 3 at point $$P.$$ The direction of electric fields are according to the charge on the sheets (away from positively charge sheet and towards the negatively charged sheet and perpendicular).

The total electric field
$$\eqalign{ & \vec E = {{\vec E}_1} + {{\vec E}_2} + {{\vec E}_3} \cr & = {E_1}( - \hat k) + {E_2}( - \hat k) + {E_3}( - \hat k) \cr & = \left[ {\frac{\sigma }{{2{\varepsilon _0}}} + \frac{{2\sigma }}{{2{\varepsilon _0}}} + \frac{\sigma }{{2{\varepsilon _0}}}} \right]( - \hat k) = - \frac{{2\sigma }}{{{\varepsilon _0}}}\hat k \cr} $$

$$\because $$ Torque on an electric dipole in an electric field,
$$\tau = p \times E \Rightarrow \left| \tau \right| = pE\sin \theta $$
where, $$\theta $$ is angle between $$E$$ and $$p$$
$$\eqalign{ & \Rightarrow 4 = p \times 2 \times {10^5} \times \sin 30 \cr & \Rightarrow p = 4 \times {10^{ - 5}}cm \cr & \Rightarrow p = q2l \cr & \therefore q2l = 4 \times {10^{ - 5}} \cr} $$
where, $$2l = 2\,cm = 2 \times {10^{ - 4}}m$$
$$\therefore q = \frac{{4 \times {{10}^{ - 5}}}}{{2 \times {{10}^{ - 2}}}} \Rightarrow 2 \times {10^{ - 3}}C = 2\,mC$$

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,
$$E = \frac{\sigma }{{K{\varepsilon _0}}}$$
Then, charge density
$$\eqalign{ & \sigma = K{\varepsilon _0}E \cr & = 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}C/{m^2} \cr} $$

$$\Delta \phi = \frac{{{\theta _{{\text{in}}}}}}{{{\varepsilon _0}}} \Rightarrow Q\,{\text{in}}\,\left( {\phi = {\phi _2} - \phi } \right){\varepsilon _0}$$

Gauss’s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by $${{\varepsilon _0}}.$$
i.e. $${\phi _{{\text{total }}}} = \frac{{{q_{{\text{inside }}}}}}{{{\varepsilon _0}}}$$
Let electric flux linked with surfaces $$A,B$$  and $$C$$ are $${\phi _A},{\phi _B}$$  and $${\phi _C}$$ respectively, i.e.
$$\eqalign{ & {\phi _{{\text{total }}}} = {\phi _A} + {\phi _B} + {\phi _C} \cr & {\text{Since,}}\,\,{\phi _C} = {\phi _A} \cr & \therefore 2{\phi _A} + {\phi _B} = {\phi _{{\text{total}}}} = \frac{q}{{{\varepsilon _0}}} \cr & {\text{or}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - {\phi _B}} \right) \cr & {\text{But}}\,\,{\phi _B} = \phi \,\,\left( {{\text{given}}} \right) \cr & {\text{Hence,}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right) \cr} $$

According to Gauss's theorem,
$$\eqalign{ & E\oint {ds} = \frac{q}{{{ \in _0}}}\,\left[ {{\text{Here}}\,E\oint {ds} = 4\pi {R^2}} \right] \cr & \therefore E = \frac{{\frac{q}{{4\pi {R^2}}}}}{{{ \in _0}}}\,\left[ {\because \frac{q}{{4\pi {R^2}}} = \sigma } \right] \cr & {\text{or}}\,\,E = \frac{\sigma }{{{\varepsilon _0}}} \cr} $$

The electric field at any point, outside or inside, the conducting sphere can depend only on $$r$$ (the radial distance from the centre of the sphere to the point).
When the two conducting spheres are connected by a conducting wire, charge will flow from one sphere (having higher potential) to other (having lower potential) till both acquire the same potential.
As, $$E = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{q}{{{r^2}}}$$
So, for different cases, $$\frac{{{E_1}}}{{{E_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = 4:1$$