q₁= 6μC
q₂ = ne = 5 x 10¹² x (1.6 x 10^-19)
q₂ = 8.0 x 10^-7 C
= 0.8 x 10^-6 C = 0.8 μC
Since electrons are removed from the sphere, q₂ is positive.
∴ Net charge on the sphere,
q = q₁ + q₂
q = (-6.0 + 0.8) x 10^-6C
q = -5.2 x 10^-6C

Given:
Total number of electrons in the slab,
N = 29 x e = 29 x 2 x 10²²
Number g electrons remvoed, n \(=\frac{q}{e}\)
\(n=\frac { 2\times { 10 }^{ -6 } }{ 1.6\times { 10 }^{ -19 } } \)
n 1.25 x 10¹³
∴ fraction of electrons removed
\(=\frac{No.of\ electrons\ removed\ (n)}{Total\ No.of \ electrons(N)}\)
\(\\ =\frac { 29\times 2\times { 10 }^{ 22 } }{ 1.25\times { 10 }^{ 13 } } =2.16\times { 10 }^{ -11 }\)

The line joining B to C is perpendicular to electric field

So potential of B = potential of C
i.e. V_B = V_c
Distance AB = 4 cm
Potential difference
between A & C = E x AB
= 5 x 10³  x (4 x 10^-2)
= 200 volt.
AC² = AB² + B²
AB² = AC² - BC²
= 25 - 9 = 16
AB = 4cm

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:
\(V=\sqrt { 2\times { 10 }^{ 4 }\times 1.76\times { 10 }^{ 11 }\times 2\times { 10 }^{ -2 } } \)
\(V=7.04\times { 10 }^{ 13 }=\sqrt { 0.704\times { 10 }^{ 14 } } \)
\(V=0.84{ 10 }^{ 7 }{ ms }^{ -1 }\)

\({ V }_{ S }-{ V }_{ R }=\frac { W }{ q } \)
Work W = 50μJ = 50 x 10^-6 J
Charge q = 2μC = 2 x 10^-6 C
V = V_S - V_R \(=\frac { W }{ q } =\frac { 50\times { 10 }^{ -6 } }{ 2\times { 10 }^{ -6 } } =25V\)
V = 25V