(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

After introduction of dielectric medium of dielectric constants K₁ and K₂ capacitor acts as if it consists of two capacitors, each having plates of area A and separation \(\frac{d}{2}\) connected in series combination for
\({ C }_{ 1 }=\frac { { \varepsilon }_{ 0 }A }{ d } \)                 ....(i)
\(\Rightarrow\) \(\frac { 1 }{ { C }_{ 2 } } =\frac { 1 }{ \left( \frac { { K }_{ 1 }{ \varepsilon }_{ 0 }A }{ d/2 } \right) } +\frac { 1 }{ \left( \frac { { K }_{ 2 }{ \varepsilon }_{ 0 }A }{ d/2 } \right) } \)
\(\Rightarrow\)\(\frac { 1 }{ { C }_{ 2 } } =\frac { 1 }{ \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) } \left( \frac { 1 }{ { 2K }_{ 1 } } +\frac { 1 }{ 2{ K }_{ 2 } } \right) \)
\(\Rightarrow\)\({ \frac { 1 }{ { C }_{ 2 } } =\frac { 1 }{ 2{ C }_{ 1 } } \left( \frac { { K }_{ 2 }+{ K }_{ 1 } }{ { K }_{ 1 }K_{ 2 } } \right) }\)
\(\Rightarrow\)\({ C }_{ 2 }={ C }_{ 1 }\left( \frac { { 2K }_{ 1 }{ K }_{ 2 } }{ { K }_{ 1 }+{ K }_{ 2 } } \right) \)

According to the diagram are when the switch S is closed, the two capacitors C₁ and C₂ in parallel will be charged by the same potential difference V.

So, charge on capacitor C₁
q₁= C₁V = 1 x 6 = 6μ
and charge on capacitor C2
q₂ = C₂V = 1 x 6 = 6μ
Hence total q = q₁ + q₂ = 6 + 6 = 12μ
When switch S is opened and dielectric is introduced. Then,

Capacity of both the capacitors becomes K times
i. e. C'₁ = C'₂ = KC = 3 x 1 = 3 μ (as C'₁= C^'₂)
Capacitor A remains connected to battery
∴ V'₁ = V = 6V
q'₁= kq₁ = 3\(\times\)6μ = 18μ
Capacitor B becomes isolated
∴ q'₂ = q₂ or C'₂V'₂ = C₂V₂ or (KC)V'₂ = CV
or \({ V }_{ 2 }=\left( \frac { V }{ K } \right) =\frac { 6 }{ 3 } =2V\)

By electrostatic induction, charge induces on the plates and opposite nature of charge appears on the surface facing each other. Therefore, they start attracting towards each other.

Field lines must be perpendicular to the plates. Also, equispaced field lines exist between two plates as electric field between them is uniform.

\(C=\frac { { \varepsilon }_{ 0 }A }{ d } \)  .......(i)
and the capacitance with metal plate is 
\({ C }^{ ' }={ \varepsilon }_{ 0 }\frac { A }{ d-d/2 } \)
\({ C }^{ ' }=\frac { 2{ \varepsilon }_{ 0 }A }{ d } \) .....(ii)
Divide (ii) by (i) 
\(\frac { { C }^{ ' } }{ C } =2\)

When a bird is perched on a bare high power line, the circuit does not get complete between the bird and the earth, therefore, nothing happens to the bird.
When a man standing on ground touches the same line, the circuit between the man and the earth gets completed. As a result, he gets a fatal shock.

Yes, in S.I. units, the capacitance of a sphere is given by C = \(4\pi\epsilon_oR\), So, \(\epsilon_o={C\over 4\pi R}\)
Clearly, SI unit of \(\epsilon_o\) is \(Fm^{-1}\)

q1 + q2 = 0, q1 = - q2. Such a system of point charges is called an an electric dipole. It is a pair of equal and opposite point charges separated by a small distance.

q₁ = +\(\mu C\), q₂ = +6\(\mu C\), F = 12N
\(q_{1}^{'}=+2-4=-2\mu C;\) \(q_2^{'}=+6-4=2\mu C\) 
F' = ?
\({F'\over F}={(q_1^{'})(q_2^{'})\over q_1q_2}={(-2)(2)\over(2)(6)}=-{1\over 3}\)
\(F'={-F\over 3}={-12\over 3}=-4 N\)(attractive)