Intensity of electric field due to a Dipole
$$E = \frac{p}{{4\pi {\varepsilon _0}{r^3}}}\sqrt {3{{\cos }^2}\theta + 1} \Rightarrow E \propto \frac{1}{{{r^3}}}$$

In equilibrium,
$$\eqalign{ & T\cos \theta = mg\,.......\left( 1 \right) \cr & T\sin \theta = qE\,.......\left( 2 \right) \cr} $$
From (1), $$T = \frac{{mg}}{{\cos \theta }}$$
From (2), $$T = \frac{{qE}}{{\sin \theta }}$$

When an electric dipole is placed in an electric field $$E,$$ a torque $$\tau = p \times E$$   acts on it. This torque tries to rotate the dipole.
If the dipole is rotated from an angle $${\theta _1}$$ to $${\theta _2}$$ then work done by external force is given by
$$W = pE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)\,......\left( {\text{i}} \right)$$
Putting $${\theta _1} = {0^ \circ },{\theta _2} = {90^ \circ }$$    in the Eq. (i), we get
$$\eqalign{ & W = pE\left( {\cos {0^ \circ } - \cos {{90}^ \circ }} \right) \cr & = pE\left( {1 - 0} \right) = pE \cr} $$

$$\eqalign{ & E = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{2p}}{{{{\left( {r - \frac{\ell }{2}} \right)}^3}}} - \frac{{2p}}{{{{\left( {r + \frac{\ell }{2}} \right)}^3}}}} \right] \cr & = \frac{1}{{4\pi { \in _0}}}\frac{{3Q}}{{{r^4}}}. \cr} $$

Let us complete the sphere. Electric field due to lower part at $$A$$ is equal to electric field due to upper part at $$B = E$$  (given) Electric field due to lower part at
$$B =$$  electric field due to full sphere - electric field due to upper part
$$\eqalign{ & = \frac{{kQ}}{{{{\left( {2R} \right)}^2}}} - E \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\rho \left( {\frac{4}{3}} \right)\pi {R^3}}}{{4{R^2}}} - E \cr & = \frac{{\rho R}}{{12{\varepsilon _0}}} - E \cr} $$

Since electric field $${\vec E}$$ decreases inside water, therefore flux $$\phi = \vec E \cdot \vec A$$   also decreases.

We have centripetal force equation
$$\eqalign{ & q\left( {\frac{{2k\lambda }}{r}} \right) = \frac{{m{v^2}}}{r} \cr & {\text{so}}\,\,v = \sqrt {\frac{{2kq\lambda }}{m}} \cr & {\text{Now,}}\,\,T = \frac{{2\pi r}}{v} = 2\pi r\sqrt {\frac{m}{{2kq\lambda }}} \cr & {\text{where}}\,\,k = \frac{1}{{4\pi {\varepsilon _0}}} \cr} $$

$$\eqalign{ & \vec E = \frac{{kQ}}{{{r^3}}}\vec r = \frac{{9 \times {{10}^9} \times 50 \times {{10}^{ - 6}}}}{{\left| {\vec r - {{\vec r}_0}} \right|}} \times \left( {\vec r - {{\vec r}_0}} \right) \cr & {\text{where}}\,\,\vec r - {{\vec r}_0} = \left( {8\hat i - 5\hat j} \right) - \left( {2\hat i + 3\hat j} \right) = 6\hat i - 8\hat j \cr & \vec E = 900\left( {3\hat i - 4\hat j} \right)V/m \cr} $$

Let us consider a spherical shell of radius $$x$$ and thickness $$dx.$$
Charge on this shell
$$dq = \rho .4\pi {x^2}dx = {\rho _0}\left( {\frac{5}{4} - \frac{x}{R}} \right).4\pi {x^2}dx$$
$$\therefore $$ Total charge in the spherical region from centre to $$r\left( {r < R} \right)$$   is
$$q = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {\frac{5}{4} - \frac{x}{R}} \right){x^2}dx} } $$

$$ = 4\pi {\rho _0}\left[ {\frac{5}{4}.\frac{{{r^3}}}{3} - \frac{1}{R}.\frac{{{r^4}}}{4}} \right] = \pi {\rho _0}{r^3}\left( {\frac{5}{3} - \frac{r}{R}} \right)$$
$$\therefore $$ Electric field at $$r,E = \frac{1}{{4\pi { \in _0}}}.\frac{q}{{{r^2}}}$$
$$ = \frac{1}{{4\pi { \in _0}}}.\frac{{\pi {\rho _0}{r^3}}}{{{r^2}}}\left( {\frac{5}{3} - \frac{r}{R}} \right) = \frac{{{\rho _0}r}}{{4{ \in _0}}}\left( {\frac{5}{3} - \frac{r}{R}} \right)$$

Considering symmetric elements each of length $$dl$$ at $$A$$ and $$B,$$ we know that electric fields perpendicular to $$PO$$ are cancelled and those along $$PO$$ are added. The electric field due to an element of length $$dl\left( { = ad\theta } \right)$$   along $$PO.$$

$$\eqalign{ & dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{d\theta }}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda dl}}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda \left( {a\,d\theta } \right)}}{{{a^2}}}\cos \theta \,\,\,\left( {\because dl = a\,d\theta } \right) \cr} $$
Net electric field at $$O$$
$$\eqalign{ & E = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} d E = 2\int_0^{\frac{\pi }{2}} {\frac{1}{{4\pi {\varepsilon _0}}}} \frac{{\lambda a\cos \theta \,d\theta }}{{{a^2}}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}}\frac{\lambda }{a}\left[ {\sin \theta } \right]_0^{\frac{\pi }{2}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a} \cdot 1 = \frac{\lambda }{{2\pi {\varepsilon _0}a}} \cr} $$