$$\phi = \vec E.\vec A = 4\hat i.\left( {2\hat i + 3\hat j} \right) = 8\,V - m$$

Electric force on charged particle is given by $$F = qE$$
Kinetic energy attained by particle = work done
$$\eqalign{ & = {\text{force}} \times {\text{ displacement}} \cr & = qE \times y \cr} $$
Alternative
Force on charged particle in a uniform electric field is $$F = ma = Eq$$
or $$a = \frac{{Eq}}{m}\,......\left( {\text{i}} \right)$$
From the equation of motion, we have
$$\eqalign{ & {v^2} = {u^2} + 2ay \cr & = 0 + 2 \times \frac{{Eq}}{m} \times y\,\,\left[ {u = 0} \right] \cr & = \frac{{2Eqy}}{m} \cr} $$
Now, kinetic energy of the particle
$$\eqalign{ & K = \frac{1}{2}m{v^2} \cr & = \frac{m}{2} \times \frac{{2Eqy}}{m} \cr & = qEy \cr} $$

$$T = PE$$  $$\sin \theta $$  Torque experienced by the dipole in an electric field, $$\overrightarrow T = \overrightarrow P \times \overrightarrow E $$
$$\eqalign{ & \overrightarrow p = p\cos \theta \hat i + p\sin \theta \hat j \cr & {\overrightarrow E _1} = E\overrightarrow i \cr & {\overrightarrow T _1} = \overrightarrow p \times {\overrightarrow E _1} = \left( {p\cos \theta \hat i + p\sin \theta \hat j} \right) \times E\left( {\hat i} \right) \cr & \tau \hat k = pE\sin \theta \left( { - \hat k} \right)\,......\left( {\text{i}} \right) \cr & {\overrightarrow E _2} = \sqrt 3 {E_1}\hat j \cr & {\overrightarrow T _2} = \left( {p\cos \theta \hat i + p\sin \theta \hat j} \right) \times \sqrt 3 {E_1}\hat j \cr & \tau \hat k = \sqrt 3 p{E_1}\cos \theta \hat k\,......\left( {{\text{ii}}} \right) \cr} $$
From eqns. (i) and (i)
$$\eqalign{ & pE\sin \theta = \sqrt 3 pE\cos \theta \cr & \tan \theta = \sqrt 3 \,\,\,\,\,\therefore \theta = {60^ \circ } \cr} $$

The electric field will be different at the location of the two charges. Therefore the two forces will be unequal. This will result in a force as well as torque.

Work done in moving the charge, $$W = Fd\,\cos \theta $$
As $$F = qE$$
$$\therefore W = qEd\cos \theta $$
or $$E = \frac{W}{{qd\cos \theta }}$$
Here, $$q = 0.2\,C, d = 2\,m$$
$$\eqalign{ & \theta = {60^ \circ },W = 4\,J \cr & \therefore E = \frac{4}{{0.2 \times 2 \times \cos {{60}^ \circ }}} \cr & = 20\,N/C \cr} $$
Alternative
As we know that potential at any point in the direction of $$\theta $$ and electric field $$E$$ is given by $$dV = - E \cdot dr\,\,\,\left( {{\text{negative sign indicates decreasing potential in direction of electric field}}} \right)$$
So, for the given situation $$dr = d\cos \theta $$
So, $$dV = Ed\cos \theta $$
Now, work done for a charge moving in potential difference $$dV$$ is given by $$W = qdV$$
$$ \Rightarrow W = qEd\cos \theta $$
Given, $$q = 0.2\,C,d = 2\,m,\theta = {60^ \circ },W = 4\,J$$
So, $$4\,J = 0.2 \times E \times 2 \times \cos {60^ \circ }$$
$$ \Rightarrow E = \frac{4}{{0.2 \times 2}} \times 2 = 20\,J$$

At equilibrium, electric force on drop balances weight of drop.
$$qE = mg \Rightarrow q = \frac{{mg}}{E} = \frac{{9.9 \times {{10}^{ - 15}} \times 10}}{{3 \times {{10}^4}}} = 3.3 \times {10^{ - 18}}C$$

Given,
Electric field $$E = 150\,N/C$$
Total surface charge carried by earth $$q = ?$$
According to Gauss’s law.
$$\eqalign{ & \phi = \frac{q}{{{ \in _0}}} = EA \cr & {\text{or,}}\,\,q = { \in _0}EA = { \in _0}E\pi {r^2} = 8.85 \times {10^{ - 12}} \times 150 \times {\left( {6.37 \times {{10}^6}} \right)^2}. \cr & \simeq 680\,KC \cr} $$
As electric field directed inward hence
$$q = - 680\,KC$$

We have $${E_a} = \frac{{2kp}}{{{r^3}}}\,{\text{and}}\,{E_e} = \frac{{kp}}{{{r^3}}};$$
$$\therefore {E_a} = 2{E_e}$$

Flux of electric field $$E$$ through any area $$A$$ is defined as $$\phi = E \cdot A\cos \theta $$    or $$\phi = E \cdot A = 0,$$    the lines are parallel to the surface.

The electric field inside a uniformly charged sphere is $$\frac{{\rho .r}}{{3{ \in _0}}}$$
The electric potential inside a uniformly charged sphere $$ = \frac{{\rho {R^2}}}{{6{ \in _0}}}\left[ {3 - \frac{{{r^2}}}{{{R^2}}}} \right]$$
$$\therefore $$ Potential difference between centre and surface
$$\eqalign{ & = \frac{{\rho {R^2}}}{{6{ \in _0}}}\left[ {3 - 2} \right] = \frac{{\rho {R^2}}}{{6{ \in _0}}} \cr & \Delta U = \frac{{q\rho {R^2}}}{{6{ \in _0}}} \cr} $$