(i) The equipotential surface is ⊥r the electric field.
(ii) The equipotential surface will be a spherical shell with the given charge at the centre

Yes, If the dielectric constant of the medium is increased, there will be a decrease of potential.

For a single charge the potential V \(=\frac { { q } }{ 4\pi { \varepsilon }_{ 0 }^{ }r } \)
This shows that V is constant if 'r' is constant. Greater the radius smaller will be the potential. In the given figure potential is increasing. This shows that the polarity of charge is positive. The direction of electric field will be radially inward. The field lines are directed from here to lower potential.

Force on particle towards upper plate B, F^-_y = qE
Vertical acceleration of particle ay=\(\frac{qE}{m}\)
Initial vertical velocity V_y = 0
Time taken by particle between the plates t = \(\frac{L}{V_x}\)
From equation of motion s = ut + \(\frac{1}{2}\) at²
Vertical deflection y = 0 + \(\frac{1}{2}\) a_yt²
\(=0+\frac { 1 }{ 2 } \left( \frac { qE }{ m } \right) { \left( \frac { L }{ { V }_{ x } } \right) }^{ 2 }\)
\(y=\frac { q{ EL }^{ 2 } }{ 2m{ V }_{ x }^{ 2 } } \)

(i) Area of the plates
(ii) separation between the plates.
(iii) Dielectric constant of the dielectric between the plates. The capacitance of a capacitor depend upon geometrical dimension and the nature of the dielectric between the plates.

The electric potential \(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { Q }{ r } \)
\(\\ V=\frac { 1 }{ r }\)
The potential due to a point charge decreases with increase of distance.
V_A - V_B > 0 ⇒ V_A > V_B
Hence the charge Q is positive.

(i) At a distance r; the force on the small sphere due to large sphere
\(F=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { Qq }{ { mr }^{ 2 } } \)
(ii) If m is the mass of small sphere then its acceleration
\(a=\frac { F }{ m } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { Qq }{ { mr }^{ 2 } } \)
(iii) As the small sphere is pushed away (i.e. r increased) 'a' decrease.
(iv) As 'a' is always +ve the speed of the small sphere goes on increasing.
(v) ∴ increasing velocity and decreasing acceleration.
(d) is correct

(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.

When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

(i) In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field.
(ii) They have a permanent dipole moment.
(iii) Examples : H₂O, N₂O, HCI and NH₃.

i) When the external electric field applied to the dielectric is very large, the bound charges (electrons) become free charges. This is called dielectric breakdown.
ii) The maximum electric field the dielectric can withstand before it breaksdown is called dielectric strength.

Electric field lines are a set of continuous lines which represent the electric field in some region of space visually.

  Coulomb force Gravitational force (i)  It is attractive or repulsive. It is always attractive . (ii) It depends on the nature of the medium in
which the two charges are kept at rest. It is independent of the medium. (iii)  It obeys Newton's third law. It does not obey Newton's Third Iaw. (iv) Coulomb force is given by,
\(\overrightarrow{F_{12}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{21}\) Gravitational force is given by,
\(\vec{F}=G \frac{m_{1} m_{2}}{r^{2}}\hat{r}\)

Coulomb's law \(\overrightarrow{F_{21}}=\frac{k q_{1} q_{2}}{r^{2}} \hat{r}_{12}\)
where, q₁ - charge; q₂ - charge
r - distance between the charges
\(\hat{r}_{12}\)- the unit vector directed from charge q₁ to charge q₂
k = Proportionality constant