(i) Dipole has two equal and opposite charges. In the uniform electric field they will experience equal and opposite force. Net force is zero, So there can't be any translatory motion
(ii) Torque \(\tau \)F = 21 sin \(\theta\) qE = p x E

V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

\(Given,radius\ of\ ring,r=2m,{ q }_{ 1 }={ 10 }^{ -6 }C\)
\({ q }_{ 2 }={ 10 }^{ -7 }C,m=9\times { 10 }^{ -4 }kg\)
\(l=2cm=0.02m,T=?\)
\(Force\ on\ particle\ is\ given\ by\)
\(F=\frac { { q }_{ 1 }{ q }_{ 2 }l }{ 4\pi { \varepsilon }_{ 0 }{ ({ l }^{ 2 }+{ r }^{ 2 }) }^{ 2 } } =\frac { (9\times { 10 }^{ 9 })({ 10 }^{ -6 }){ 10 }^{ -7 })l }{ ({ 4+0.0004 })^{ 3/2 } } \)
\(=\frac { 9\times { 10 }^{ -4 }l }{ 8.001 } =1.125\times { 10 }^{ -4 }l\)
\(As,acceleration,a=\frac { F }{ m } =\frac { 1.125\times { 10 }^{ -4 }l }{ 9{ \times 10 }^{ -4 } } \)
\(Thus,the\ motion\ of\ particle\ is\ SHM.\)
\(Now,\ thime\ period\ of\ oscillation\ of\ particle\ is\quad given\ by\)
\(T=2\pi \sqrt { \frac { l }{ a } } =2\pi \sqrt { \frac { l }{ 0.125l } } \Rightarrow 17.7s\)

Electric flux \(\phi\) is a measure of number of field lines crossing a surface. The number of field lines passing through unit area (N / S) will be proportional to the electric field, i.e. N / S \(\propto \) E \(\Rightarrow\) N \(\propto \) ES  
The quantity ES is the electric flux through surface S. As in the given question, the field lines that enter the closed surface leave the surface immediately, so the net electric flux is bound to the system. Thus, electric flux is zero.