Charge and Charge Density (Distribution of Charges) Questions in English

(C) The surface charge density $\sigma$ is defined as the charge $q$ per unit area $A$ of the surface.
Mathematically,it is expressed as $\sigma = \frac{q}{A}$.
From this relation,it is clear that $\sigma$ is directly proportional to the charge $q$ and inversely proportional to the surface area $A$.

(A) The surface of a charged conductor is an equipotential surface because there is no potential difference between any two points on it. Therefore,the potential at point $Q$ is the same as at point $P$,i.e.,$V_Q = V_P = V$.
The surface charge density $\sigma$ is inversely proportional to the radius of curvature $r$ $(\sigma \propto 1/r)$. Since the radius of curvature at $Q$ is larger than at $P$,the charge density at $Q$ is less than at $P$ $(\sigma_Q < \sigma_P = \sigma)$.
The electric field strength $E$ at the surface of a conductor is given by $E = \sigma / \epsilon_0$. Since $\sigma_Q < \sigma_P$,the electric field at $Q$ is less than at $P$ $(E_Q < E_P = E)$.
Thus,at point $Q$,the values are: charge density $< \sigma$,potential $= V$,and electric field $< E$.

(A) For a system of parallel conducting plates,the charge on the outermost surfaces is equal to half the sum of the total charges on all plates.
Let the total charge on the system be $Q_{total} = Q_A + Q_B + Q_C = Q + 0 + (-2Q) = -Q$.
The charge on the outermost surfaces (left surface of $A$ and right surface $S$ of $C$) is given by $q_{outer} = Q_{total} / 2 = -Q / 2$.
Thus,the charge appearing on the outer surface $S$ of plate $C$ is $-Q / 2$.

(D) For a system of parallel conducting plates,the charge on the outermost surfaces is equal to half the sum of the total charges on all plates.
Let the total charge on the plates be $Q_{total} = Q + 0 + (-2Q) = -Q$.
The charge on the outermost surfaces (leftmost surface of the first plate and rightmost surface of the third plate) is given by $q_{outer} = Q_{total} / 2 = -Q/2$.
Thus,the charge on the outer surface of the rightmost plate is $-Q/2$.

(B) The upper half of the ring has a positive linear charge density $\lambda$. The electric field at the centre $O$ due to this positive charge will be directed away from the arc,i.e.,towards the bottom-right direction.
The lower half of the ring has a negative linear charge density $-\lambda$. The electric field at the centre $O$ due to this negative charge will be directed towards the arc,i.e.,also towards the bottom-right direction.
By symmetry,the horizontal components of the electric fields from both halves cancel out,while the vertical components add up in the downward direction. However,looking at the geometry of the arc placed on the left of the centre $O$,the net electric field vector points towards the right,specifically along the direction $OB$.

(B) The total charge $Q$ is given by the integral of the volume charge density over the volume of the sphere:
$Q = \int_0^R \rho(r) \cdot 4\pi r^2 dr$
Substituting $\rho(r) = \frac{A}{r^2} e^{-2r/a}$:
$Q = \int_0^R \left( \frac{A}{r^2} e^{-2r/a} \right) (4\pi r^2) dr = 4\pi A \int_0^R e^{-2r/a} dr$
Evaluating the integral:
$Q = 4\pi A \left[ \frac{e^{-2r/a}}{-2/a} \right]_0^R = 4\pi A \left( -\frac{a}{2} \right) (e^{-2R/a} - e^0)$
$Q = -2\pi aA (e^{-2R/a} - 1) = 2\pi aA (1 - e^{-2R/a})$
Rearranging to solve for $R$:
$1 - e^{-2R/a} = \frac{Q}{2\pi aA}$
$e^{-2R/a} = 1 - \frac{Q}{2\pi aA}$
Taking the natural logarithm on both sides:
$-\frac{2R}{a} = \log \left( 1 - \frac{Q}{2\pi aA} \right)$
$R = -\frac{a}{2} \log \left( 1 - \frac{Q}{2\pi aA} \right) = \frac{a}{2} \log \left( \frac{1}{1 - \frac{Q}{2\pi aA}} \right)$

(C) The total positive charge $Q_{in}$ on the inner solid sphere is calculated by integrating the volume charge density over the volume of the sphere:
$Q_{in} = \int_{0}^{R_1} \rho(r) \cdot 4\pi r^2 dr = \int_{0}^{R_1} \frac{\rho_0}{r} \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_{0}^{R_1} r dr = 4\pi \rho_0 \left[ \frac{r^2}{2} \right]_0^{R_1} = 2\pi \rho_0 R_1^2$.
The total negative charge $Q_{out}$ on the outer hollow sphere is given by the surface charge density multiplied by its surface area:
$Q_{out} = -\sigma \cdot 4\pi R_2^2 = -4\pi \sigma R_2^2$.
Given that the total charge in the system is zero, we have $Q_{in} + Q_{out} = 0$, which implies $Q_{in} = |Q_{out}|$.
$2\pi \rho_0 R_1^2 = 4\pi \sigma R_2^2$.
Rearranging for the ratio $\frac{R_2^2}{R_1^2}$:
$\frac{R_2^2}{R_1^2} = \frac{2\pi \rho_0}{4\pi \sigma} = \frac{\rho_0}{2\sigma}$.
Taking the square root of both sides, we get:
$\frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}}$.

(A) The square plate has side length $a$ and is centered at the origin $(0,0)$. The limits for $x$ and $y$ are from $-a/2$ to $a/2$.
The total charge $Q$ is given by the surface integral of the charge density $\sigma$ over the area of the plate:
$Q = \int \sigma dA = \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} (xy) dx dy$
This can be separated into two independent integrals:
$Q = \left( \int_{-a/2}^{a/2} x dx \right) \left( \int_{-a/2}^{a/2} y dy \right)$
Since the integrand $x$ is an odd function and the limits are symmetric about the origin,$\int_{-a/2}^{a/2} x dx = 0$.
Similarly,$\int_{-a/2}^{a/2} y dy = 0$.
Therefore,$Q = 0 \times 0 = 0$.
The total charge on the plate is zero.

(D) Let the charges on the shells be $Q_A = Q$,$Q_B = 2Q$,and $Q_C = -Q$.
$1$. The inner surface of shell $A$ has $0$ charge,so the outer surface of $A$ has $Q$.
$2$. The inner surface of shell $B$ must have a charge of $-Q$ to cancel the effect of the charge on shell $A$. Since the total charge on shell $B$ is $2Q$,the outer surface of $B$ has $2Q - (-Q) = 3Q$.
$3$. The inner surface of shell $C$ must have a charge of $-3Q$ to cancel the effect of the charge on the outer surface of $B$. Since the total charge on shell $C$ is $-Q$,let the charge on the outer surface of $C$ be $q_{out}$.
$4$. Thus,$-3Q + q_{out} = -Q$,which gives $q_{out} = 2Q$.
$5$. The ratio of the charge on the inner surface to the charge on the outer surface of shell $C$ is $\frac{-3Q}{2Q} = -\frac{3}{2}$.

(A) The surface charge density $\sigma$ is defined as the total charge $Q$ divided by the total surface area $A$ of the object.
For a cube with side length $a = 5 \, cm = 0.05 \, m$,the total surface area is $A = 6a^2$.
$A = 6 \times (0.05 \, m)^2 = 6 \times 25 \times 10^{-4} \, m^2 = 150 \times 10^{-4} \, m^2 = 0.015 \, m^2$.
The total charge is $Q = 6 \, \mu C = 6 \times 10^{-6} \, C$.
The surface charge density is $\sigma = \frac{Q}{A} = \frac{6 \times 10^{-6}}{6 \times 25 \times 10^{-4}} \, C/m^2$.
$\sigma = \frac{10^{-6}}{25 \times 10^{-4}} \, C/m^2 = \frac{1}{25} \times 10^{-2} \, C/m^2 = 0.04 \times 10^{-2} \, C/m^2 = 4 \times 10^{-4} \, C/m^2$.
Converting to $\mu C/m^2$: $\sigma = 4 \times 10^{-4} \times 10^6 \, \mu C/m^2 = 4 \times 10^2 \, \mu C/m^2$.

(D) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively. Given $q_1 + q_2 = Q$.
Since surface charge densities $\sigma$ are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.
This implies $\frac{q_1}{q_2} = \frac{r^2}{R^2}$.
Using the ratio,we find $q_1 = \frac{Qr^2}{R^2 + r^2}$ and $q_2 = \frac{QR^2}{R^2 + r^2}$.
The potential at the common centre is $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.
Substituting the values of $q_1$ and $q_2$,we get $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Qr}{R^2 + r^2} + \frac{QR}{R^2 + r^2} \right) = \frac{Q(R + r)}{4\pi\varepsilon_0(R^2 + r^2)}$.

(D) Given: Radius $(R) = 10 \, cm$,Potential $(V) = 300 \, V$.
In $CGS$ units,$1 \, statvolt = 300 \, V$. Therefore,the potential $V = 1 \, statvolt$.
The potential of a charged sphere is given by $V = \frac{Q}{R}$.
Substituting the values: $1 = \frac{Q}{10}$,which gives $Q = 10 \, statcoulomb$.
The surface charge density $(\sigma)$ is given by $\sigma = \frac{Q}{4 \pi R^2}$.
Substituting the values: $\sigma = \frac{10}{4 \pi (10)^2} = \frac{10}{400 \pi} = \frac{1}{40 \pi} \approx 0.00796 \approx 8 \times 10^{-3} \, CGS \, esu/cm^2$.

(D) The total charge $Q$ on the plate is given by the surface integral of the charge density $\sigma$ over the area of the plate:
$Q = \int_{-a}^{a} \int_{-a}^{a} \sigma_0 xy \, dx \, dy$
$Q = \sigma_0 \left( \int_{-a}^{a} x \, dx \right) \left( \int_{-a}^{a} y \, dy \right)$
Since the integral of an odd function over a symmetric interval $[-a, a]$ is zero:
$\int_{-a}^{a} x \, dx = 0$ and $\int_{-a}^{a} y \, dy = 0$
Therefore,$Q = \sigma_0 \times 0 \times 0 = 0$.
Alternatively,by symmetry,the charge in the first quadrant is positive,the second is negative,the third is positive,and the fourth is negative. Due to the symmetry of the function $\sigma = \sigma_0 xy$ about the axes,the total charge sums to zero.

(D) The charge on a small element of the ring of angular width $d\theta$ is given by $dq = \lambda \cdot dl$,where $dl = a \, d\theta$.
Given $\lambda = {\lambda _0} \cos \theta$,we have $dq = (\lambda _0 \cos \theta) (a \, d\theta)$.
The semicircular ring spans from $\theta = 0$ to $\theta = \pi$.
Therefore,the total charge $Q$ is:
$Q = \int_{0}^{\pi} \lambda _0 a \cos \theta \, d\theta$
$Q = \lambda _0 a \int_{0}^{\pi} \cos \theta \, d\theta$
$Q = \lambda _0 a [\sin \theta]_{0}^{\pi}$
$Q = \lambda _0 a (\sin \pi - \sin 0)$
$Q = \lambda _0 a (0 - 0) = 0$.
Thus,the total charge on the ring is zero.

(B) For a charged conductor in electrostatic equilibrium,the entire conductor is an equipotential surface,meaning the electric potential is the same at every point.
However,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $R$ $(\sigma \propto 1/R)$.
At the sharpest point,the radius of curvature is minimum,which makes the surface charge density maximum.
Since the electric field near the surface is given by $E = \sigma / \epsilon_0$,the electric field is also maximum at the sharpest point.

(D) The square plate has side length $a$ and is centered at the origin $(0,0)$ in the $xy$-plane. The limits for $x$ and $y$ are from $-\frac{a}{2}$ to $+\frac{a}{2}$.
The total charge $Q$ is given by the surface integral of the charge density $\sigma$ over the area of the plate:
$Q = \iint \sigma \, dA = \int_{-\frac{a}{2}}^{\frac{a}{2}} \int_{-\frac{a}{2}}^{\frac{a}{2}} (xy) \, dx \, dy$
This can be separated into two independent integrals:
$Q = \left( \int_{-\frac{a}{2}}^{\frac{a}{2}} x \, dx \right) \left( \int_{-\frac{a}{2}}^{\frac{a}{2}} y \, dy \right)$
Since the integrand $x$ is an odd function and the limits are symmetric about the origin,$\int_{-\frac{a}{2}}^{\frac{a}{2}} x \, dx = 0$.
Similarly,$\int_{-\frac{a}{2}}^{\frac{a}{2}} y \, dy = 0$.
Therefore,$Q = 0 \times 0 = 0$.
The total charge on the plate is zero.

(N/A) Consider a long thin wire $XY$ of uniform linear charge density $\lambda$.
Consider a point $A$ at a perpendicular distance $l$ from the mid-point $O$ of the wire.
Let $E$ be the electric field at point $A$ due to the wire $XY$.
Consider a small length element $dx$ on the wire section at a distance $x$ from $O$ (i.e.,$OZ = x$).
The charge on this element is $dq = \lambda dx$.
The electric field due to this element at point $A$ is:
$dE = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda dx}{(AZ)^{2}}$
Since $AZ = \sqrt{l^{2} + x^{2}}$,we have:
$dE = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda dx}{l^{2} + x^{2}}$
The electric field is resolved into two rectangular components. $dE \cos \theta$ is the perpendicular component and $dE \sin \theta$ is the parallel component. When the whole wire is considered,the parallel components $dE \sin \theta$ cancel out due to symmetry. Only the perpendicular component $dE \cos \theta$ contributes to the net electric field at point $A$.
Thus,the effective electric field $dE_{1}$ is:
$dE_{1} = dE \cos \theta = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda dx \cos \theta}{l^{2} + x^{2}} \dots (1)$
In $\Delta AZO$,$\tan \theta = \frac{x}{l} \Rightarrow x = l \tan \theta$. Differentiating,$dx = l \sec^{2} \theta d\theta \dots (2)$
Also,$l^{2} + x^{2} = l^{2} + l^{2} \tan^{2} \theta = l^{2} \sec^{2} \theta \dots (3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$dE_{1} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\lambda (l \sec^{2} \theta d\theta) \cos \theta}{l^{2} \sec^{2} \theta} = \frac{\lambda}{4 \pi \epsilon_{0} l} \cos \theta d\theta$
For an infinitely long wire,$\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Integrating:
$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4 \pi \epsilon_{0} l} \cos \theta d\theta = \frac{\lambda}{4 \pi \epsilon_{0} l} [\sin \theta]_{-\pi/2}^{\pi/2}$
$E = \frac{\lambda}{4 \pi \epsilon_{0} l} [1 - (-1)] = \frac{2\lambda}{4 \pi \epsilon_{0} l} = \frac{\lambda}{2 \pi \epsilon_{0} l}$

(N/A) The electric potential $V$ at a point $P$ due to a continuous linear charge distribution with linear charge density $\lambda$ along a line element $dl$ is given by the integral:
$V = \frac{1}{4\pi\epsilon_0} \int \frac{\lambda dl}{r}$
where:
$1. \epsilon_0$ is the permittivity of free space.
$2. \lambda$ is the linear charge density (charge per unit length).
$3. dl$ is the infinitesimal length element of the distribution.
$4. r$ is the distance from the charge element $dl$ to the point $P$ where the potential is being calculated.

Linear charge density $(\lambda):$ The amount of charge per unit length is called linear charge density $(\lambda)$. $\lambda = \frac{Q}{l}$,where $Q$ is the total charge and $l$ is the length.
Unit: $C/m$. Dimensional formula: $[M^0 L^{-1} T^1 A^1]$.
Surface charge density $(\sigma):$ The amount of charge per unit area is called surface charge density $(\sigma)$. $\sigma = \frac{Q}{A}$,where $Q$ is the total charge and $A$ is the area.
Unit: $C/m^2$. Dimensional formula: $[M^0 L^{-2} T^1 A^1]$.
Volume charge density $(\rho):$ The amount of charge per unit volume is called volume charge density $(\rho)$. $\rho = \frac{Q}{V}$,where $Q$ is the total charge and $V$ is the volume.
Unit: $C/m^3$. Dimensional formula: $[M^0 L^{-3} T^1 A^1]$.

(N/A) $(1)$ Line charge distribution: Suppose a line is divided into smaller elements of length $dl$. Let $\vec{r}$ be the position vector of a small element with linear charge density $\lambda$,so its charge is $dq = \lambda dl$.
Consider a point $P$ with position vector $\vec{R}$. Let $r^{\prime}$ be the distance from the element $dl$ to $P$,and $\hat{r}^{\prime}$ be the unit vector pointing from the element to $P$. The electric field at $P$ due to the element is:
$\vec{dE} = \frac{k \lambda dl}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{l} \frac{k \lambda dl}{(r^{\prime})^{2}} \hat{r}^{\prime}$
$(2)$ Surface charge distribution: Suppose a surface $S$ is divided into small elements $\Delta S$. Let $\sigma$ be the surface charge density,so the charge on an element is $dq = \sigma dS$.
The electric field at point $P$ due to the surface element is:
$\vec{dE} = \frac{k \sigma dS}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{S} \frac{k \sigma dS}{(r^{\prime})^{2}} \hat{r}^{\prime}$
$(3)$ Volume charge distribution: Suppose a volume $V$ has a charge density $\rho$. The charge in a small volume element $dV$ is $dq = \rho dV$.
The electric field at point $P$ due to the volume element is:
$\vec{dE} = \frac{k \rho dV}{(r^{\prime})^{2}} \hat{r}^{\prime}$
By the superposition principle,the total electric field at $P$ is:
$\vec{E} = \int_{V} \frac{k \rho dV}{(r^{\prime})^{2}} \hat{r}^{\prime}$

(N/A) Charge distribution is the way electric charge is spread over a region. It is classified into three types:
$1$. Linear Charge Distribution: When charge is distributed uniformly along a line (like a thin wire or ring),it is called linear charge distribution. The linear charge density $\lambda$ is defined as $\lambda = \frac{dq}{dl}$,where $dq$ is the charge on an element of length $dl$. Its $SI$ unit is $C/m$.
$2$. Surface Charge Distribution: When charge is distributed uniformly over a surface (like a thin sheet or shell),it is called surface charge distribution. The surface charge density $\sigma$ is defined as $\sigma = \frac{dq}{dA}$,where $dq$ is the charge on an area element $dA$. Its $SI$ unit is $C/m^2$.
$3$. Volume Charge Distribution: When charge is distributed uniformly throughout the volume of an object,it is called volume charge distribution. The volume charge density $\rho$ is defined as $\rho = \frac{dq}{dV}$,where $dq$ is the charge in a volume element $dV$. Its $SI$ unit is $C/m^3$.

(N/A) $1$. Linear Charge Density $(\lambda)$: It is defined as the charge per unit length of a conductor. $\lambda = \frac{dq}{dl}$. Its $SI$ unit is $C/m$.
$2$. Surface Charge Density $(\sigma)$: It is defined as the charge per unit surface area of a conductor. $\sigma = \frac{dq}{dA}$. Its $SI$ unit is $C/m^2$.
$3$. Volume Charge Density $(\rho)$: It is defined as the charge per unit volume of a conductor. $\rho = \frac{dq}{dV}$. Its $SI$ unit is $C/m^3$.

(A) Volume charge density $\rho$ is defined as the charge per unit volume.
Mathematically,$\rho = \frac{dq}{dV}$.
For a small volume element $\Delta V$,the charge $\Delta q$ contained within it is given by the product of the volume charge density and the volume element.
Therefore,$\Delta q = \rho \Delta V$.

(A) The electric field $E$ due to an infinite line charge with linear charge density $\lambda$ at a radial distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.
Given $\lambda = 8 \times 10^{-9} \, C/m$ and $r = x = 3 \, m$.
The electric field at $x = 3 \, m$ is $E = \frac{8 \times 10^{-9}}{2 \pi \varepsilon_0 (3)}$.
Using the boundary condition for the surface charge density $\sigma$ at a conductor surface,$E = \frac{\sigma}{\varepsilon_0}$,we have $\sigma = E \varepsilon_0$.
Substituting $E$,we get $\sigma = \frac{\lambda}{2 \pi r} = \frac{8 \times 10^{-9}}{2 \pi (3)}$.
$\sigma = \frac{8 \times 10^{-9}}{6 \pi} \approx 0.424 \times 10^{-9} \, C/m^2 = 0.424 \, nC/m^2$.

(D) The potential $V$ at the center of a spherical shell due to a surface charge distribution is given by the integral $V = \oint \frac{k dq}{R}$.
Since the radius $R$ is constant for all points on the surface of the sphere,we can take $\frac{k}{R}$ outside the integral.
$V = \frac{k}{R} \oint dq$.
The integral $\oint dq$ represents the total charge $Q$ on the sphere.
Therefore,$V = \frac{kQ}{R}$.
Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V = \frac{Q}{4 \pi \varepsilon_0 R}$.
Note that the potential at the center of a charged spherical shell depends only on the total charge $Q$ and the radius $R$,and is independent of the distribution of the charge on the surface.

(C) For a conducting sphere, any excess charge given to it resides entirely on its outer surface due to the electrostatic property of conductors.
Since both the solid and hollow conducting spheres have the same radius $R$, their surface areas are identical $(A = 4\pi R^2)$.
Because the charge distribution is restricted to the surface in both cases, the maximum charge that can be held before dielectric breakdown of the surrounding medium occurs is the same for both spheres.

(B) The surface charge density $\sigma$ is defined as the charge per unit area,given by $\sigma = \frac{Q}{A} = \frac{Q}{4 \pi r^2}$.
Since the charges $Q$ on both spheres are equal,we have $\sigma \propto \frac{1}{r^2}$.
Given radii are $r_1 = 2 \,cm$ and $r_2 = 4 \,cm$.
The ratio of charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{r_2^2}{r_1^2}$.
Substituting the values,$\frac{\sigma_1}{\sigma_2} = \frac{(4)^2}{(2)^2} = \frac{16}{4} = \frac{4}{1}$.
Thus,the ratio is $4: 1$.

(B) When a charge $Q$ is given to one of two large parallel conducting plates,the charge redistributes itself on both sides of each plate such that the electric field inside the conducting material is zero.
Let the charges on the four surfaces (from left to right) be $q_1, q_2, q_3,$ and $q_4$.
For the first plate: $q_1 + q_2 = Q$ and for the second plate: $q_3 + q_4 = 0$.
Since the plates are large,the electric field due to a surface with charge density $\sigma$ is $E = \frac{\sigma}{2 \varepsilon_0} = \frac{q}{2 A \varepsilon_0}$.
Inside the conducting material of the plates,the net electric field must be zero.
Applying this condition,we find that the charges on the inner surfaces are $q_2 = \frac{Q}{2}$ and $q_3 = -\frac{Q}{2}$,while the outer surfaces have $q_1 = \frac{Q}{2}$ and $q_4 = \frac{Q}{2}$.
The electric field at a point between the plates is due to the inner surface charges $q_2$ and $q_3$.
$E_{\text{net}} = E_2 + E_3 = \frac{q_2}{2 A \varepsilon_0} + \frac{|q_3|}{2 A \varepsilon_0} = \frac{Q/2}{2 A \varepsilon_0} + \frac{Q/2}{2 A \varepsilon_0} = \frac{Q}{4 A \varepsilon_0} + \frac{Q}{4 A \varepsilon_0} = \frac{Q}{2 A \varepsilon_0}$.

(B) Let the surface charge densities on the shells be $\sigma_1, \sigma_2, \sigma_3$ respectively.
Given that $\sigma_1 = \sigma_2 = \sigma_3 = \sigma$.
The surface charge density is defined as $\sigma = \frac{Q}{A}$,where $A = 4\pi r^2$ is the surface area of the sphere.
For the first shell: $Q_1 = \sigma \cdot 4\pi R^2$.
For the second shell: $Q_2 = \sigma \cdot 4\pi (2R)^2 = \sigma \cdot 16\pi R^2$.
For the third shell: $Q_3 = \sigma \cdot 4\pi (3R)^2 = \sigma \cdot 36\pi R^2$.
Now,the ratio $Q_1 : Q_2 : Q_3$ is:
$Q_1 : Q_2 : Q_3 = (\sigma \cdot 4\pi R^2) : (\sigma \cdot 16\pi R^2) : (\sigma \cdot 36\pi R^2)$.
Dividing by $4\pi R^2$,we get:
$Q_1 : Q_2 : Q_3 = 1 : 4 : 9$.

(C) Given the equation for electric flux: $\phi = \alpha \sigma + \beta \lambda$.
By the principle of homogeneity of dimensions,the dimensions of each term must be equal: $[\phi] = [\alpha \sigma] = [\beta \lambda]$.
From $[\phi] = [\alpha \sigma]$,we get $[\alpha] = \frac{[\phi]}{[\sigma]}$.
From $[\phi] = [\beta \lambda]$,we get $[\beta] = \frac{[\phi]}{[\lambda]}$.
Now,consider the ratio $\frac{\alpha}{\beta}$: $\left[\frac{\alpha}{\beta}\right] = \frac{[\phi]/[\sigma]}{[\phi]/[\lambda]} = \frac{[\lambda]}{[\sigma]}$.
The dimensions of linear charge density $\lambda$ are $[Q/L]$ and surface charge density $\sigma$ are $[Q/L^2]$.
Substituting these: $\left[\frac{\alpha}{\beta}\right] = \frac{[Q/L]}{[Q/L^2]} = \frac{L^2}{L} = [L]$.
Since the dimension is $[L]$,the ratio $\left(\frac{\alpha}{\beta}\right)$ represents length,which is a unit of displacement.

(C) For a charged conductor,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $(ROC)$ at that point,i.e.,$\sigma \propto \frac{1}{ROC}$.
From the figure,the radius of curvature at the sharpest point (top) is the smallest,and it increases as we move towards the flatter sides.
Comparing the points:
$1$. The point at $\sigma_1$ has the smallest radius of curvature.
$2$. The point at $\sigma_3$ has a larger radius of curvature than $\sigma_1$ but smaller than the sides.
$3$. The points at $\sigma_2$ and $\sigma_4$ are on the flatter sides,where the radius of curvature is the largest and equal.
Thus,the order of radius of curvature is $(ROC)_1 < (ROC)_3 < (ROC)_2 = (ROC)_4$.
Since $\sigma \propto \frac{1}{ROC}$,the order of surface charge densities is $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$.
Therefore,the correct option is $(C)$.

(A) The total charge $q_{\text{total}}$ on the disc is calculated by integrating the surface charge density over the area of the disc.
Consider a small elemental ring of radius $r$ and width $dr$. The area of this ring is $ds = 2 \pi r dr$.
The total charge is given by:
$q_{\text{total}} = \int_0^R \sigma ds$
Substituting $\sigma = \sigma_0 r^3$ and $ds = 2 \pi r dr$:
$q_{\text{total}} = \int_0^R (\sigma_0 r^3)(2 \pi r dr)$
$q_{\text{total}} = 2 \pi \sigma_0 \int_0^R r^4 dr$
Performing the integration:
$q_{\text{total}} = 2 \pi \sigma_0 \left[ \frac{r^5}{5} \right]_0^R$
$q_{\text{total}} = \frac{2 \pi \sigma_0 R^5}{5}$

(B) The electric field $E$ at a distance $r$ from the center of a charged conducting sphere of radius $R$ is given by Gauss's Law as $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}$.
Since the total charge $Q$ on the sphere is related to the surface charge density $\sigma$ by $Q = \sigma \cdot (4 \pi R^{2})$,we substitute this into the electric field equation.
$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{\sigma (4 \pi R^{2})}{r^{2}}$.
Simplifying the expression,we get $E = \frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}$.
Rearranging the formula to solve for $\sigma$,we get $\sigma = E \varepsilon_{0} \left(\frac{r}{R}\right)^{2}$.

(A) $1$. According to the principle of electrostatic induction,when a charge $-q$ is placed at the center of a conducting shell,an equal and opposite charge $+q$ is induced on the inner surface of the shell to ensure the electric field inside the conductor is zero.
$2$. The inner surface of the shell now has a charge of $+q$. The surface charge density on the inner surface is given by $\sigma_{inner} = \frac{q}{4 \pi r_1^2}$.
$3$. Since the shell has a total charge $Q$ and the inner surface has a charge $+q$,the charge on the outer surface must be $Q_{outer} = Q - q$ to conserve the total charge.
$4$. The surface charge density on the outer surface is given by $\sigma_{outer} = \frac{Q-q}{4 \pi r_2^2}$.
$5$. Therefore,the surface charge densities are $\frac{q}{4 \pi r_1^2}$ and $\frac{Q-q}{4 \pi r_2^2}$.

(D) As shown in the figure,consider an elementary charge $dq$ having length $dx$ at a distance $x$ from the point charge $q$. The linear charge density is $\lambda = \frac{Q}{L}$.
Then,$dq = \lambda dx = \frac{Q}{L} dx$.
By Coulomb's law,the force $dF$ between the elementary charge $dq$ and the point charge $q$ is:
$dF = \frac{k q dq}{x^2} = \frac{k q Q dx}{L x^2}$.
To find the total force $F$,integrate $dF$ from $x = r$ to $x = r + L$:
$F = \int_r^{r+L} \frac{k Q q}{L x^2} dx = \frac{k Q q}{L} \int_r^{r+L} x^{-2} dx$.
$F = \frac{k Q q}{L} \left[ -\frac{1}{x} \right]_r^{r+L} = \frac{k Q q}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right)$.
$F = \frac{k Q q}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right) = \frac{k Q q}{L} \left( \frac{r+L-r}{r(r+L)} \right)$.
$F = \frac{k Q q}{L} \left( \frac{L}{r(r+L)} \right) = \frac{k Q q}{r(r+L)}$.

(A) As shown in the figure,let us consider an element of angular width $d \theta$ at an angle $\theta$.
The length of this element is $dl = a d \theta$.
The charge on this element is $dq = \lambda dl = (\lambda_0 \cos^2 \theta) (a d \theta)$.
The total charge $Q$ on the circle is obtained by integrating $dq$ over the entire circumference from $0$ to $2 \pi$:
$Q = \int_{0}^{2 \pi} \lambda_0 \cos^2 \theta a d \theta$
$Q = a \lambda_0 \int_{0}^{2 \pi} \cos^2 \theta d \theta$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$:
$Q = a \lambda_0 \int_{0}^{2 \pi} \frac{1 + \cos 2 \theta}{2} d \theta$
$Q = \frac{a \lambda_0}{2} [\theta + \frac{\sin 2 \theta}{2}]_{0}^{2 \pi}$
$Q = \frac{a \lambda_0}{2} [(2 \pi + 0) - (0 + 0)]$
$Q = \frac{a \lambda_0}{2} (2 \pi) = \pi a \lambda_0$
Thus,the total charge is $\pi a \lambda_0$.

(B) The surface charge density $\sigma$ is defined as the charge $Q$ per unit area $A$ of the surface.
Given: Radius $r = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.
Surface charge density $\sigma = \frac{3}{\pi} \ \mu C/m^2$.
The surface area of a spherical shell is $A = 4 \pi r^2$.
Using the formula $\sigma = \frac{Q}{A}$,we get $Q = \sigma \times A = \sigma \times 4 \pi r^2$.
Substituting the values:
$Q = \left( \frac{3}{\pi} \ \mu C/m^2 \right) \times 4 \pi \times (0.25 \ m)^2$
$Q = 3 \times 4 \times (0.0625) \ \mu C$
$Q = 12 \times 0.0625 \ \mu C = 0.75 \ \mu C$.
Therefore,the charge required is $0.75 \ \mu C$.

(B) Given: Diameter of the sphere $d = 2.4 \, m$.
Radius of the sphere $r = \frac{d}{2} = 1.2 \, m$.
Surface charge density $\sigma = 80.0 \, \mu C m^{-2} = 80 \times 10^{-6} \, C m^{-2}$.
The total charge $Q$ on the surface of the sphere is given by the product of surface charge density and surface area $A$.
$Q = \sigma \times A = \sigma \times (4 \pi r^2)$.
Substituting the values:
$Q = 80 \times 10^{-6} \times 4 \times 3.14159 \times (1.2)^2$.
$Q = 80 \times 10^{-6} \times 4 \times 3.14159 \times 1.44$.
$Q \approx 1.4476 \times 10^{-3} \, C$.
Rounding to the nearest value, $Q \approx 1.45 \times 10^{-3} \, C$.

(A) Given: Mass of the particle $m = 2 \times 10^{-6} \ kg$,charge on the particle $q = 5 \times 10^{-6} \ C$.
The electric field $E$ due to a charged conducting surface is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.
For the particle to hang in the air,the electric force must balance the gravitational force acting on it.
$F_e = F_g \Rightarrow qE = mg$
Substituting $E = \frac{\sigma}{\epsilon_0}$ into the equation:
$q \left( \frac{\sigma}{\epsilon_0} \right) = mg$
$\sigma = \frac{mg \epsilon_0}{q}$
Substituting the given values:
$\sigma = \frac{(2 \times 10^{-6} \ kg) \times (10 \ m \ s^{-2}) \times (8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2})}{5 \times 10^{-6} \ C}$
$\sigma = \frac{20 \times 8.85 \times 10^{-18}}{5 \times 10^{-6}}$
$\sigma = 4 \times 8.85 \times 10^{-12} \ C \ m^{-2}$
$\sigma = 35.4 \times 10^{-12} \ C \ m^{-2}$

(B) Let the point charge $q$ be at the origin $x=0$. The wire extends from $x = 2 \text{ cm}$ to $x = 12 \text{ cm}$.
The linear charge density of the wire is $\lambda = \frac{Q}{L} = \frac{24 \times 10^{-6} \text{ C}}{0.1 \text{ m}} = 2.4 \times 10^{-4} \text{ C/m}$.
Consider a small element $dx$ of the wire at a distance $x$ from the point charge $q$. The charge on this element is $dq = \lambda dx$.
The force between the point charge $q$ and the element $dq$ is $dF = \frac{k q dq}{x^2} = \frac{k q \lambda dx}{x^2}$,where $k = 9 \times 10^9 \text{ N} \cdot \text{m}^2 / \text{C}^2$.
The total force $F$ is the integral of $dF$ from $x = 2 \text{ cm} = 0.02 \text{ m}$ to $x = 12 \text{ cm} = 0.12 \text{ m}$:
$F = \int_{0.02}^{0.12} \frac{k q \lambda}{x^2} dx = k q \lambda \left[ -\frac{1}{x} \right]_{0.02}^{0.12} = k q \lambda \left( \frac{1}{0.02} - \frac{1}{0.12} \right)$
$F = (9 \times 10^9) \times (1 \times 10^{-6}) \times (2.4 \times 10^{-4}) \times \left( 50 - 8.333 \right) = 9000 \times 2.4 \times 10^{-4} \times \left( \frac{6-1}{0.12} \right) = 9000 \times 2.4 \times 10^{-4} \times \frac{5}{0.12} = 90 \text{ N}$.