(i) q₁ = q₂ = 6.5 x 10^-7C
r = 50 cm = 0.5m
Electrostatic force of repulsion,
\(F=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
\(\\ =\frac { 9\times { 10 }^{ 9 }\times { (6.5\times { 10 }^{ -7 }) }^{ 2 } }{ { (0.5) }^{ 2 } } \)
F = 1.521 x 10^-2N
(ii) Now if q₁,q₂ are doubled and r' is halved then F becomes 16 times.
i.e., New force of repulsion, \(F'=16\times \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
F' = 16F
= 16 x 1.521510^-2
F' = 0.24 N.

(i) The equivalent capacitance is,
C_p = C₁+ C₂ + C₃
= (1 + 2 + 3) = 6μF
(ii) Total charge, q = CμV
= 6 x 10^-6 x 100 = 600μC
q₁ = C₁V = 1 x 100 = 100μC
q₂ = C₂V = 2 x 100 = 200μC
q₃ = C₃V = 3 x 100 = 300μC

Area of the plate, A = 25 cm²
= 25 x 10^-4m²
Distance between the plates, d = 2 mm
d = 2 x 10^-3m
Potential difference, V = 12V
Charge, q = CV \(\left[ C=\frac { { \varepsilon }_{ 0 }A }{ d } \right] \)
\(q=\left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) V\)
\(=\frac { 8.85\times { 10 }^{ -12 }\times 25\times { 10 }^{ -4 }\times 12 }{ 2\times { 10s }^{ -3 } } \)
q = 1.33 x 10^-10C

Capacitance, C = 10μF = 10 x 10^-6F
Voltage, V = 10V
Energy, E = ?
Energy stored in the capacitor, E = \(\frac{1}{23}CV^2\)
= \(\frac{1}{2}\) x 10 x 10^-6 x 10 x 10
= 5 x 10^-4 J

Given: Electric field strength E = 10⁴ NC^-1 between the plates
Distance of separation between the plates = 2 cm
= 2 x 10^-2 m.
Velocity of the electron when it reaches B = V = ?
To find:
According to equation of motion V = u² + 2as
u - initial velocity = 0; a \(=\frac { F }{ m } =\frac { Ee }{ m } =E\left( \frac { e }{ m } \right) \)
Formula: V² = 2as
Solution:
\(V=\sqrt { 2\times { 10 }^{ 4 }\times 1.76\times { 10 }^{ 11 }\times 2\times { 10 }^{ -2 } } \)
\(V=7.04\times { 10 }^{ 13 }=\sqrt { 0.704\times { 10 }^{ 14 } } \)
\(V=0.84{ 10 }^{ 7 }{ ms }^{ -1 }\)