\({ \varepsilon }_{ r }=\frac { \varepsilon }{ { \varepsilon }_{ 0 } } =80\)
ε = 80 x ε₀
= 80 x 8.854 x 10^-12
= 708 x 10^-12
ε₀ = 7.08 x 10^-10 C²N^-1m^-2.

The total positive charge in the nucleus is,
q = +ze = 79 x 1.6 x 10^-19C.
Volume charge density, \(\sigma =\frac { q }{ \frac { 4 }{ 3 } \times \pi { R }^{ 3 } } \)
\(s=\frac { q }{ \frac { 4 }{ 3 } \times 3.14\times (7.5\times { 10 }^{ -15 }{ ) }^{ 3 } } \)
= 0.088 x 1026
= 8.8 x 10²⁴ Cm^-3

r = 25cm = 25 x 10^-2m
\(\sigma =\frac { 3 }{ \pi } { cm }^{ -2 }\)
\(AS,\quad \sigma =\frac { q }{ A } =\frac { q }{ 4\pi { r }^{ 2 } } \) [A - surface Area of the sphere]
\(q=(4\pi { r }^{ 2 })\sigma \)
\(=4\pi (0.25{ ) }^{ 2 }\times \frac { 3 }{ \pi } \)
= 0.75C.

For charge -q to be in equilibrium, for an -q due to +Q at point A should be equal and opposite to that due to +Q at point B.
i.e \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Qq }{ { x }^{ 2 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q.q }{ { (r-x) }^{ 2 } } \)
r (r - x)²
x r - x
x = \(\frac{r}{2}\)
∴ for equilibrium, -q must be kept at the middle of the line joining the A and B.

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

Charge of an electron, e = 1.6 x 10^-19C
q = 1C
∴ No of electrons, n = \(\frac{q}{e}\)
\(n=\frac { 1 }{ 1.6\times { 10 }^{ -19 } } \)
= 6.25 x 10¹⁸
n = 6.25 x 10¹⁸ electrons

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
\(=\frac{No.of\ electrons\ removed\ (n)}{Total\ No.of \ electrons(N)}\)
\(\\ =\frac { 29\times 2\times { 10 }^{ 22 } }{ 1.25\times { 10 }^{ 13 } } =2.16\times { 10 }^{ -11 }\)

Here, q = -3 x 10^-7 C
Charge of one electron, e = -1.6 x 10^-19 C
Number of electrons transferred from wool to
polythenepiece, n = \(\frac{q}{e}\)\(=\frac { -3\times { 10 }^{ -7 }C }{ -1.6\times { 10 }^{ -19 }C } \)
= 1.875 x 10¹²

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

(i) Electric flux \(\phi =\frac { Total\ enclosed\ charge }{ { \varepsilon }_{ 0 } } \)
Net charge enclosed inside the shell q = 0
∴ electric flux through the shell \(\frac { q }{ { \varepsilon }_{ 0 } } =0\)
(ii) The electric field or net charge inside the spherical conducting shell is zero.
Hence the force on charge \(\frac{Q}{2}\)is zero.
Force on charge at \(A,{ F }_{ A }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2Q\left( Q+\frac { Q }{ 2 } \right) }{ { x }^{ 2 } } \)

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

The system of charges has zero electric potential energy.
The system of charges shown in figure has zero electric potential energy.
\(U=\frac { { q }^{ 2 } }{ a } -K.\frac { { q }^{ 2 } }{ 2a } -K.\frac { { q }^{ 2 } }{ 2a } =0\)
U = 0

\({ V }_{ S }-{ V }_{ R }=\frac { W }{ q } \)
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

(i) Consider a positive charge q kept fixed at the origin which produces an electric field \(\overset { \rightarrow }{ E } \) around it.
(ii) A positive test charge q' is brought from point R to point P against the repulsive force between q and q' as shown in Figure. Work must be done to overcome this repulsion. This work done is stored as potential energy.
(iii) The test charge q' is brought from R to P with constant velocity which means that external force used to bring the test charge q' from R to P must be equal and opposite to the coulomb force \(\left( { \overset { \rightarrow }{ E } }_{ ext }=-{ \overset { \rightarrow }{ F } }_{ coloumb } \right) \) 
The work done is
\(W=\int _{ R }^{ P }{ { \overset { \rightarrow }{ F } }_{ ext } } .d\overset { \rightarrow }{ r } \quad \quad \quad ...(1)\)
(iii) Since coulomb force is conservative, work done is independent of the path and it depends only on the initial and final positions of the test charge. If potential energy associated with q' at P is Up and that at R is UR' then difference in potential energy is defined as the work done to bring a test charge q' from point P to R and is given as U_p - U _R = W.
\(\Delta U=\int _{ R }^{ P }{ { \overset { \rightarrow }{ F } }_{ ext } } .d\overset { \rightarrow }{ r } \)
\(Since{ \overset { \rightarrow }{ F } }_{ ext }=-{ \overset { \rightarrow }{ F } }_{ coloumb }=-q'\overset { \rightarrow }{ E } \)
\(\Delta U=\int _{ R }^{ P }{ \left( -q'\overset { \rightarrow }{ E } \right) } .d\overset { \rightarrow }{ r } =q'\int _{ R }^{ P }{ \left( -\overset { \rightarrow }{ E } \right) } .d\overset { \rightarrow }{ r } \)

(i) Consider the following charged object of irregular shape as shown in Figure. The entire charged object is divided into a large number of charge elements \(\Delta { q }_{ 1 },\Delta { q }_{ 2 },\Delta { q }_{ 3 },.....\Delta { q }_{ n },\). and each charge element \(\Delta { q }\) is taken as a point charge.
(ii) The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { { \Delta q }_{ 1 } }{ { r }_{ 1p }^{ 2 } } { \hat { r } }_{ 1p }+\frac { \Delta { q }_{ 2 } }{ { r }_{ 2p }^{ 2 } } { \hat { r } }_{ 2p }+.....+\frac { \Delta { q }_{ n } }{ { r }_{ np }^{ 2 } } { \hat { r } }_{ np } \right) \)
\(\approx \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \sum _{ t=1 }^{ n }{ \frac { { \Delta q }_{ 1 } }{ { r }_{ 1P }^{ 2 } } { \hat { r } }_{ iP } } (1)\)
(iii) Here \(\Delta { q }_{ i }\) is the i^th charge element, r_ip is the unit vector from the i^th charge element to the point P. However the equation (1) is only an approximation. To incorporate the continuous distribution of charge, we take the limit \(\Delta q\rightarrow 0(=dq).\) In this limit, the summation in the equation (1) becomes an integration and takes the following form \(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \frac { dq }{ { r }^{ 2 } } \hat { r } } \)   ...(2)
(iv) Here r is the distance of the point P from the infinitesimal charge dq and \(\hat { r } \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution can be difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } =q\vec { E } \)

(a) If the charge Q is uniformly distributed along the line of length L, then linear charge density (charge per unit length) \(\lambda \) is \(\lambda =\frac { Q }{ L } \) unit is coulomb per meter (Cm^-1). The charge present in the infinitesimal length dl is dq =\(\lambda \)dl. This is shown in Figure 1(a).
The electric field due to the line of total charge Q is given by
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \frac { \lambda dl }{ { r }^{ 2 } } } \hat { r } =\frac { \lambda }{ 4\pi { \varepsilon }_{ 0 } } \int { \frac { dl }{ { r }^{ 2 } } } \hat { r } \)
(b) If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) \(\sigma \) is \(\sigma \) =\(\frac{Q}{A}\). Its unit is coulomb per square meter (C m^-2). The charge present in the infinitesimal area dA is dq=\(\sigma dA\). This is shown in the figure 1(b). The electric field due to a total charge Q is given by \(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \frac { \sigma da }{ { r }^{ 2 } } \hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \sigma \int { \frac { da }{ { r }^{ 2 } } } \hat { r } } \) This is shown in Figure 1(b).
(c) If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) p is given by \(\rho =\frac { Q }{ V } .\) Its unit is coulomb per cubic meter (Cm^-3).
The charge present in the infinitesimal volume element dV is dq = pdV. This is shown in Figure 1(c). The electric field due to a volume of total charge Q is given by,\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \frac { \rho dV }{ { r }^{ 2 } } =\hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \rho \int { \frac { dV }{ { r }^{ 2 } } \hat { r } } } \).