No

Zero

ML^-1T^-2

0

In the dielectric medium between the plates.

On touching, charges neutralize. Therefore,
F = 0

In a satellite, there is a condition of weightlessness. Therefore, mg = 0. On account of electrostatic force of repulsion between the balls, the strings would become horizontal. Therefore, angle between the strings = 180^o.
Also, tension in each string = force of repulsion 
\(T={1\over 4\pi\epsilon_o}{q^2\over(2l)^2}N\)

We know that dielectric constant of a medium is
 \(K=\epsilon_r={\epsilon\over\epsilon_o}\)
\(\epsilon=K\epsilon_o=1\times8.854\times10^{-12}\)
= \(8.854\times10^{-12}C^2N^{-1}m^{-2}\)

(i) Like charges repel and unlike charges attract each other.
(ii) Charge is quantized
(iii) Charge is conserved 
(iv) Charge on a body is not affected by its motion.

Total number of electrons in the sphere = \(29\times2\times10^{22}\)
No of electrons removed = \({q\over e}={2\times10^{-6}\over1.6\times10^{-19}}=1.25\times10^{13}\)
Fraction of electrons removed = \({1.25 \times10^{13}\over 29\times2\times10^{23}}=2.16\times10^{-11}\)

Additivity of charge means the total charge on a system is the algebraic sum (with proper signs) of all individual charges in the system.

Charge on any body or particle can be integral multiple of charge on an electron (-e), i.e.,
q = \(\pm ne\), where n = 1, 2, 3,....

The ebonite rod acquires a negative charge and fur/ wool acquires an equal positive charge.

Its mass decreases slightly as it losses some electrons.

The cause of charging is actual transfer of electrons from one body to the other.

\(3\mu F\)

(i) The equivalent capacitance of C₁ and C₂ connected in series
\(\frac { 1 }{ C' } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } \)
\(\Rightarrow C'=\frac { 6 }{ 2n } =3\mu F\)
∴ Charge, q' = C'V = (3μF)12 = 36μF
∴ Charge on each capadtor of C₁ and C₂ is 36 μC
∴ Charge on C₃,
q₃ = C₃V = (6μF) x 12 = 72μC
q₃ = 72μC
(ii) Equivalent capacitance of network,
\({ C }_{ eq }=\frac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } +{ C }_{ 3 }\)
\(=\frac { 6\times 6 }{ 6+6 } +6\)
= 3 + 6 = 9μF
C_eq = 9μF
(iii) Energy stored in the network of capacitors
U = U₁ + U₂ + U₃ = \(\frac { { q' }^{ 2 } }{ { 2C }_{ 1 } } +\frac { { q }'^{ 2 } }{ { 2C }_{ 2 } } +\frac { { q }^{ 2 } }{ { 2C }_{ 3 } } \)
∵ C₁ = C₂ = C₃ = 6μF
\(\therefore U=\frac { 1 }{ (12\mu F) } [{ q' }^{ 2 }+{ q' }^{ 2 }+{ q }^{ 2 }]\)
\(=\frac { 1 }{ (12\mu F) } [({ 36\mu C) }^{ 2 }+({ 36\mu C) }^{ 2 }+({ 72\mu C) }^{ 2 }]\)
\(U=648\mu J\)