Yes, two balls having same kind of charge can attract each other if charge possessed by one ball is very large as compared to that on the other ball because when two charged balls are placed near each other, they induced opposite charges on the faces of each other. On the ball having small amount of charge, very large amount of induced charge is produced, two balls get attracted towards each other.

\(m=3.2\times { 10 }^{ -27 }kg,\) 
\({ q }=1.6\times { 10 }^{ -19 }C,\ g=10{ ms }^{ -2 }\) 
F = mg
F = qE
qE = mg
\(E=\frac { mg }{ q } \) 
\(\frac { 3.2\times { 10 }^{ -27 }\times 10 }{ 1.6\times { 10 }^{ -19 } } \) 
\(E=2.0\times { 10 }^{ -7 }{ NC }^{ -1 }\) 

\(Given\quad r=3.2\times { 10 }^{ -15 }m,\) 
\({ q }_{ 1 }={ q }_{ 2 }=2\times 1.6\times { 10 }^{ -19 }C\) 
\(F=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \) 
\(\frac { 9\times { 10 }^{ 9 }\times { \left( 2\times 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ { \left( 3.2\times { 10 }^{ -15 } \right) }^{ 2 } } \) 
\(=90N\) 

\(E=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2pr }{ { \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } } \) 
\(E=\frac { 9\times { 10 }^{ 9 }\times 2\times q\times 2a\times r }{ { \left( { r }^{ 2 }-{ a }^{ 2 } \right) }^{ 2 } } \) 
\(=\frac { 9\times { 10 }^{ 9 }\times 2\times 5\times { 10 }^{ -6 }\times 4\times { 10 }^{ -2 }\times 4\times { 10 }^{ -2 } }{ { \left( 16\times { 10 }^{ -4 }-4\times { 10 }^{ -21 } \right) }^{ 2 } } \) 
\(or\quad E=\frac { 144 }{ 144\times { 10 }^{ -8 } } ={ 10 }^{ 8 }{ NC }^{ -1 }\)

Zero, because equal forces are inclined at angle of \(120°\) so resultant is Zero.

(a) We know that the net field inside a charged conductor is zero i.e. 
\(\overrightarrow { E } \) = 0, inside
Let us choose a gaussian surface lying wholly within the conductor and enclosing the cavity.
According to Gauss' law,
\(\oint \overrightarrow { E } .\overrightarrow { dS } =\frac { q }{ { \varepsilon }_{ 0 } } =0\)  \((\because \overrightarrow { E } =0,inside)\)
\(\therefore \) q = 0 i.e. charge inside the cavity is zero. Hence the entire charge Q on the conductor must appear on the outer surface of the conductor.
(b) The conductor B carrying a charge + q inseted in the cavity induces a charge - q on the metal surface of cavity and + q on the outside surface of the conductor A [Fig (b)]. As the outer surface of A originally had a charge Q, the total charge on it would become (Q + q).
(c) To shift a sensitive instrument from the strong electrostatic fields in its environment, enclose the instrument fully by a metallic surface.

Since \({ F }_{ e }=-\frac { { ke }^{ 2 } }{ { r }^{ 2 } } \)
So     \({ ke }^{ 2 }={ -F }_{ e }{ r }^{ 2 }\)
And    \({ F }_{ G }=-G\frac { { m }_{ p }{ m }_{ e } }{ { r }^{ 2 } } \)
So \({ Gm }_{ p }{ m }_{ e }={ -F }_{ G }{ r }^{ 2 }\)
So dimensions of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } =\frac { { F }_{ e }{ r }^{ 2 } }{ { F }_{ G }{ r }^{ 2 } } =\frac { { F }_{ e } }{ { F }_{ G } } \)
 \(=\frac { { [MLT }^{ -2 }] }{ { [MLT }^{ -2 }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]\)
 = No dimensions
The value of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } \)
  \(=\frac { 9\times { 10 }^{ 9 }\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 6.67\times { 10 }^{ -11 }(1.67\times { 10 }^{ -27 })(9.1\times { 10 }^{ -31 }) } \)
  \(=2.9\times { 10 }^{ 39 }\)
From Eq. (1) and (2), we have
  \(\left| \frac { { F }_{ e } }{ { F }_{ G } } \right| =\frac { { ke }^{ 2 } }{ { Gm }_{ p }{ m }_{ e } } =2.9\times { 10 }^{ 39 }\)
The ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

Given q₁ = 0.4 \(\mu\)C = 0.4 \(\times\) 10^-6C
 q₂ = -0.8 \(\mu\)C = -0.8 \(\times\) 10^-6C
 F = 0.2 N, r = ?
(a) Since \(F=9\times { 10 }^{ 9 }\times \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
or  \(0.2=\frac { 9\times { 10 }^{ 9 }\times (0.4\times { 10 }^{ -6 })(0.8\times { 10 }^{ -6 }) }{ { r }^{ 2 } } \)
or  \({ r }^{ 2 }=\frac { 9\times { 10 }^{ 9 }\times 0.32\times { 10 }^{ -12 } }{ 0.2 } \)
or \(r=12\times { 10 }^{ -2 }m=12cm\)
(b) The second sphere will attract the first with the same force i.e. force of 0.2 N.

Given, capscitance, C = 2 F
and separation between plates,
d = 0.5 cm = 0.5 \(\times\) 10^-2 m
Capscitance of a parallel plate capacitor, C=\(\frac{\varepsilon_{0}A}{d}\)
or A=\(\frac{Cd}{\varepsilon_{0}}=\frac{2\times 0.5\times 10^{-2}}{8.854\times 10^{-12}}\)
= 1.13 \(\times\)10⁹ m² = 1130 km²
This area is very large, so it is not possible that the capacitance of a capacitor is too large as 2F. So, the capacitance of any capacitor should be the range of 2 \(mu\)F.

Given, C₁ = C₂ = 600 pF
= 600 \(\times\)10^-12F
= 6 \(\times\)10^-10 F
V₁ = 200 V, V₂ = 0
\(\begin{aligned} \therefore \text { Energy lost } & =\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)} \\ \end{aligned}\)
\(\begin{aligned} =\frac{\left(6 \times 10^{-10}\right)^2(200-0)^2}{2 \times 12 \times 10^{-10}} \end{aligned}\)
= 6 \(\times\)10^-6 J

Given,
Capacitance, C = 8pF.
In the first case, the parallel plates are at a distance ‘d’  and is filled with air.
Air has dielectric constant, k = 1
Capacitance, C\(=\frac{k \times \epsilon_{o} \times A}{d}=\frac{\epsilon_{o} \times A}{d}\)          .......(i)
Here,
A = area of each plate
ϵ_o​ = permittivity of free space.
Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2
Given, dielectric constant of the substance, k₁ = 6
Hence, the capacitance of the capacitor,
\(\mathrm{C}_{1}=\frac{k_{1} \times \epsilon_{o} \times A}{d_{1}}=\frac{6 \epsilon_{0} \times A}{d / 2}=\frac{12 \epsilon_{o} A}{d}\)         .......(ii)
Taking ratios of eqns. (1) and (2), we get,
C₁ = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.
Hence, the capacitance between the plates is 96pF.

Potential difference, V = 15 x 10⁶ V
Dielectric strength of the surrounding gas = 5 x 10⁷ V/m
Electric field intensity, E = Dielectric strength = 5 x 10⁷ V/m
Minimum radius of the spherical shell required for the purpose is given by,
\(r=\frac{V}{E}\)
\(=\frac{15 \times 10^{6}}{5 \times 10^{7}}=0.3 \mathrm{~m}=30 \mathrm{~cm}\)
Hence, the minimum radius of the spherical shell required is 30 cm.

Electric field of 1st spherical conductor on its surface
\({ E }_{ 1 }\propto \frac { 1 }{ a } \)
Electric field of 2nd spherical conductor on its surface
\({ E }_{ 2 }\propto \frac { 1 }{ b }\)
\(\frac { { E }_{ 1 } }{ { E }_{ 2 } } =\frac { b }{ a } \) 
Since a flat portion may be considered as a spherical surface of large radius (i.e. lower charge density, as charge density \(=\left( \frac { charge }{ area } \right) \) and a pointed portion as of small radius (i.e. highest charge density).

Given,
Capacitance of the capacitor, C = 12pF = 12 x 10^-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
\(\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 12 \times 10^{-12} \mathrm{\times}(50)^{2} \mathrm{~J}=1.5 \times 10^{-8} \mathrm{~J}\)
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8 J. was disconnected.

There are three capacitors cach of capacitance 9 pF.
\(\therefore\) C₁ = C₂ = C₃ = 9 pF
and voltage, V = 120 V
(i) The total capcitance in series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
\(\Rightarrow \frac{1}{C_{s}}=\frac{3}{9} \Rightarrow C_{s}=3 pF\)
(ii) Let the charge across the system be q and potentials across C₁, C₂ and C₃ be V₁, V₂ and V₃, respectively.
Charge, q = C_s. V = 3 \(\times\)120 = 360 pC
Potential difference across C₁,
\(V_{1}=\frac{q}{C_{1}}=\frac{360}{9}=40 V\)
Potential difference across C₂,
\(V_{2}=\frac{q}{C_{2}}=\frac{360}{9}=40 V\)
Potential difference across C₃,
\(V_{3}=\frac{q}{C_{3}}=\frac{360}{9}=40 V\)
Thus, the potential difference across each capacitor is 40 V.