(i) In electric field line is the path of movement of a positive test charge (q₀ ⇾ 0) A moving charge experiences a continuous force is an electric field, so field line is always continuous
(ii) The field lines nevel intersect since if they cross, there will be two directions of electric field at the point of intersection, which is impossible.

(i) Electric flux through one face \(=\frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \mu { V }_{ m }\)
(ii) By symmetry the flux through each of the six faces of cube will be same when charge is placed at the centre.
\(\\ \therefore { \phi }_{ E }=\frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)
Thus electric flux passing through two opposite faces of the cube
\(=2\times \frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)
\(\phi =\frac { 1 }{ 3 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)

(i) Charge on each sphere \(=\frac { { q }_{ A }-{ 3q }_{ A } }{ 2 } =-{ q }_{ A }\)
(ii)New charge on A is \(\frac { { q }_{ A } }{ 2 } \)
New charge on B is \(\frac { { q }_{ A }(2{ q }_{ B }) }{ 4 } \)
\(\because \frac { \frac { { q }_{ A } }{ 2 } +{ q }_{ B } }{ 2 } =\frac { { q }_{ A } }{ 4 } +\frac { q_{ B } }{ 2 } =\frac { { q }_{ A }+2{ q }_{ B } }{ 4 } \)
q_B = -3q_A
∵ q_B = -3q_A
∴ New charge on B₁ is \(\frac { { q }_{ A }-{ 6q }_{ A } }{ 4 } \)
New charge on B \(-\frac { 5 }{ 4 } { q }_{ A }\)

Capacitor is a device used to store electric charge and electrical energy.

When an external electric field is applied, the dipoles inside the polar molecule tend to align in the direction of the electric field. Hence a net dipole moment is induced in it. Then the dielectric is said to be polarized by an external electric field.