Electric Field and usage of Gauss's Law Questions in English

(D) Let $\phi_A, \phi_B,$ and $\phi_C$ be the electric flux linked with the surfaces $A, B,$ and $C$ respectively.
According to Gauss's Law,the total electric flux through the closed surface is $\phi_{total} = \phi_A + \phi_B + \phi_C = \frac{q}{\varepsilon_0}$.
Due to the symmetry of the cylinder,the flux linked with the two plane surfaces $A$ and $C$ is equal,so $\phi_A = \phi_C$.
Substituting this into the Gauss's Law equation,we get $2\phi_A + \phi_B = \frac{q}{\varepsilon_0}$.
Given that the flux through the curved surface $B$ is $\phi_B = \phi$,we have $2\phi_A + \phi = \frac{q}{\varepsilon_0}$.
Rearranging for $\phi_A$,we get $2\phi_A = \frac{q}{\varepsilon_0} - \phi$.
Therefore,$\phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

(D) For a charged conducting spherical shell of radius $R$,the electric field at a distance $r$ from the centre is given by Gauss's Law.
For $r > R$,the shell acts as a point charge,so $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.
For $r < R$,the electric field inside a charged conducting shell is always zero because there is no enclosed charge within a Gaussian surface drawn inside the conductor.
Since the distance $\frac{R}{2}$ is less than $R$,the electric field at this point is $0$.

(B) According to Gauss's Law,the total electric flux through a closed surface is $\phi_{total} = \frac{q_{enclosed}}{\varepsilon_0}$.
When a point charge $q$ is placed at one corner of a cube,it is shared by $8$ identical cubes to enclose the charge completely in a symmetric manner.
Therefore,the flux through one cube is $\frac{1}{8}$ of the total flux through the larger symmetric enclosure.
Thus,the electric flux passing through the given cube is $\phi = \frac{1}{8} \left( \frac{q}{\varepsilon_0} \right) = \frac{q}{8\varepsilon_0}$.

(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \oint E \cdot dA = \frac{q_{en}}{\varepsilon_0}$.
Given the electric field $E = Ar$ is acting radially outward,for a spherical surface of radius $a$,the electric field at the surface is $E = Aa$.
The surface area of the sphere is $S = 4\pi a^2$.
Since the field is uniform over the surface of the sphere and directed radially outward,the flux is $\phi = E \times S = (Aa) \times (4\pi a^2) = 4\pi A a^3$.
Equating this to Gauss's Law: $4\pi A a^3 = \frac{q}{\varepsilon_0}$.
Therefore,the charge contained is $q = 4\pi \varepsilon_0 A a^3$.

(C) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2\varepsilon_0}$.
For a uniform electric field,the potential difference $V$ between two points separated by a distance $d$ is given by $V = E \times d$.
Substituting the expression for $E$,we get $V = \frac{\sigma}{2\varepsilon_0} \times d$.
Rearranging for $d$,we have $d = \frac{V \times 2\varepsilon_0}{\sigma}$.
Given: $V = 50 \, V$,$\sigma = 0.10 \, \mu C/m^2 = 0.10 \times 10^{-6} \, C/m^2$,and $\varepsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)$.
Substituting the values: $d = \frac{50 \times 2 \times 8.85 \times 10^{-12}}{0.10 \times 10^{-6}}$.
$d = \frac{100 \times 8.85 \times 10^{-12}}{10^{-7}} = 8.85 \times 10^{-3} \, m$.
$d = 8.85 \, mm$.

(B) The electric field $E$ due to an infinite nonconducting sheet of charge is given by $E = \frac{\sigma}{2\varepsilon_0}$.
The potential difference $V$ between two points separated by a distance $d$ in a uniform electric field is $V = E \cdot d$.
Substituting the expression for $E$,we get $V = \frac{\sigma}{2\varepsilon_0} \cdot d$.
Rearranging for $d$,we have $d = \frac{2\varepsilon_0 V}{\sigma}$.
Given $\sigma = 10^{-7} \ C/m^2$,$V = 5 \ V$,and $\varepsilon_0 = 8.854 \times 10^{-12} \ C^2/(N \cdot m^2)$.
$d = \frac{2 \times 8.854 \times 10^{-12} \times 5}{10^{-7}} = 88.54 \times 10^{-5} \ m = 0.8854 \times 10^{-3} \ m = 0.88 \ mm$.

(B) $1$. According to the property of conductors in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.
$2$. By Gauss's Law,if we consider a Gaussian surface inside the conducting material,the net charge enclosed must be zero.
$3$. $A$ point charge $+Q$ is placed at the center. To make the net charge inside the conducting material zero,an induced charge of $-Q$ must appear on the inner surface of the shell (at radius $a$).
$4$. The total charge on the shell is given as $-q$. Let the charge on the outer surface be $q_{outer}$.
$5$. Since the total charge on the shell is the sum of the charges on the inner and outer surfaces,we have: $-Q + q_{outer} = -q$.
$6$. Solving for $q_{outer}$,we get: $q_{outer} = -q + Q$.
$7$. Therefore,the charge distribution is $-Q$ on the inner surface and $-q + Q$ on the outer surface.

(A) According to the shell theorem,the electric field inside a uniformly charged spherical shell is zero. Therefore,the electric force on the charge $q$ due to the shell in which it is placed (let's call it shell $A$) is zero.
The electric force on the charge $q$ is only due to the other shell (shell $B$). The distance between the center of shell $B$ and the charge $q$ is $10d + d/2 = 19d/2$.
Since shell $B$ acts as a point charge $Q$ at its center for external points,the force is given by Coulomb's Law:
$F = \frac{1}{4\pi \varepsilon_0} \frac{Qq}{(19d/2)^2} = \frac{Qq}{4\pi \varepsilon_0 (361d^2/4)} = \frac{Qq}{361\pi \varepsilon_0 d^2}$.
Since both charges are positive,the force is repulsive,meaning it acts away from shell $B$,which is towards the left.

(D) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.
For a sheet with charge density $+\sigma$,the field is directed away from the sheet. For a sheet with charge density $-\sigma$,the field is directed towards the sheet.
Outside the plates,the fields due to the two sheets are equal in magnitude and opposite in direction,so the net electric field is $E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.
Between the plates,the fields due to both sheets are directed from the positive sheet to the negative sheet (towards the right). Thus,the net electric field is $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.
Therefore,the statements in $A$ and $C$ are correct.

(D) Since the electric field converges at the origin,the electric field vector $\vec{E}$ is directed towards the origin. Thus,$\vec{E} = -100\,r\hat{r}$.
By Gauss's Law,$\oint \vec{E} \cdot d\vec{s} = \frac{q_{enclosed}}{\epsilon_0}$.
For a spherical surface of radius $r$ centered at the origin,the flux is $\Phi = E \cdot (4\pi r^2) = (-100r)(4\pi r^2) = -400\pi r^3$.
Since the flux is negative,the enclosed charge $q$ must be negative,which confirms that the charge contained in any spherical volume centered at the origin is negative.
For $r = 3\,cm = 0.03\,m$,the charge $q = \epsilon_0 \Phi = (8.854 \times 10^{-12}) \times (-400\pi \times (0.03)^3) \approx -3 \times 10^{-13}\,C$.
Thus,all the given statements are correct.

(D) When two large parallel conducting plates are placed close to each other,the charges on the outer faces are equal to half the sum of the total charges on the plates. Let the charges on the four faces be $q_1, q_2, q_3$,and $q_4$ from left to right.
$q_1 = q_4 = \frac{Q + 0}{2} = \frac{Q}{2}$
Since plate $X$ has total charge $Q$,$q_1 + q_2 = Q \implies q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$.
Since plate $Y$ has total charge $0$,$q_3 + q_4 = 0 \implies q_3 = -q_4 = -\frac{Q}{2}$.
The electric field due to a single charged sheet is $E = \frac{\sigma}{2\varepsilon_0} = \frac{q}{2A\varepsilon_0}$.
At point $A$ (to the left of both plates): The fields from all four faces point to the left. $E_A = \frac{q_1 + q_2 + q_3 + q_4}{2A\varepsilon_0} = \frac{Q/2 + Q/2 - Q/2 + Q/2}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the left).
At point $B$ (between the plates): The fields from $q_1, q_2$ point to the right,and from $q_3, q_4$ point to the left. $E_B = \frac{q_1 + q_2 - q_3 - q_4}{2A\varepsilon_0} = \frac{Q/2 + Q/2 - (-Q/2) - Q/2}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the right).
At point $C$ (to the right of both plates): The fields from all four faces point to the right. $E_C = \frac{q_1 + q_2 + q_3 + q_4}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the right).
Thus,the magnitude of the electric field at $A, B$,and $C$ is $\frac{Q}{2A\varepsilon_0}$. Therefore,all statements are correct.

(B) According to Gauss's Law,for a thin spherical shell of radius $R$ with a uniformly distributed charge $Q$:
$1$. Inside the shell $(r < R)$: The electric field $E$ is zero because there is no enclosed charge.
$2$. Outside the shell $(r \ge R)$: The shell acts as a point charge located at its center,so the electric field is given by $E = k\frac{Q}{r^2}$,where $k = \frac{1}{4\pi\epsilon_0}$.
Thus,the graph of $E(r)$ versus $r$ should show $E = 0$ for $0 \le r < R$ and a $1/r^2$ decay for $r \ge R$. This corresponds to the graph shown in option $B$.

(C) To find the electric field at a distance $r_1$ from the center,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.
For a spherical Gaussian surface of radius $r_1$,$E(4\pi r_1^2) = \frac{q_{enc}}{\varepsilon_0}$.
The charge enclosed $q_{enc}$ is the integral of the charge density $\rho(r)$ over the volume of the sphere of radius $r_1$:
$q_{enc} = \int_0^{r_1} \rho(r) (4\pi r^2) dr = \int_0^{r_1} \left( \frac{Q}{\pi R^4} r \right) (4\pi r^2) dr = \frac{4Q}{R^4} \int_0^{r_1} r^3 dr = \frac{4Q}{R^4} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4}$.
Substituting this into Gauss's Law:
$E(4\pi r_1^2) = \frac{Q r_1^4}{\varepsilon_0 R^4} \implies E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

(C) Consider a spherical shell of radius $x$ and thickness $dx$. The charge on this shell is given by $dq = \rho(x) \cdot 4 \pi x^2 dx = \rho_0 \left( \frac{5}{4} - \frac{x}{R} \right) \cdot 4 \pi x^2 dx$.
The total charge $q$ enclosed within a sphere of radius $r$ $(r < R)$ is:
$q = \int_0^r dq = 4 \pi \rho_0 \int_0^r \left( \frac{5}{4} x^2 - \frac{x^3}{R} \right) dx$
$q = 4 \pi \rho_0 \left[ \frac{5}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right] = 4 \pi \rho_0 \left( \frac{5 r^3}{12} - \frac{r^4}{4R} \right) = \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right)$.
Using Gauss's Law,the electric field $E$ at distance $r$ is:
$E \cdot 4 \pi r^2 = \frac{q}{\varepsilon_0}$
$E = \frac{1}{4 \pi \varepsilon_0 r^2} \cdot \left[ \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right) \right]$
$E = \frac{\rho_0 r}{4 \varepsilon_0} \left( \frac{5}{3} - \frac{r}{R} \right)$.

(D) The electric field $E$ is related to the potential $\phi$ by the relation $E = -\frac{d\phi}{dr}$.
Given $\phi = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 + b) = -2ar$.
According to Gauss's Law in differential form,the charge density $\rho$ is given by $\nabla \cdot E = \frac{\rho}{\varepsilon_0}$.
In spherical coordinates,for a radially symmetric field $E(r)$,the divergence is $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{\rho}{\varepsilon_0}$.
Substituting $E = -2ar$ into the equation:
$\frac{\rho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.
Therefore,the charge density is $\rho = -6a\varepsilon_0$.

(B) For a uniformly charged sphere of total charge $Q$ and radius $R$:
$1$. Inside the sphere $(r < R)$: The electric field is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q r}{R^3}$. This shows that $E \propto r$,meaning the electric field increases linearly from the center $(r=0)$ to the surface $(r=R)$.
$2$. At the surface $(r = R)$: The electric field is maximum,$E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}$.
$3$. Outside the sphere $(r > R)$: The sphere behaves as a point charge at the center,so $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$. This shows that $E \propto \frac{1}{r^2}$,meaning the electric field decreases inversely with the square of the distance.
Comparing this behavior with the given options,the graph that shows a linear increase for $r < R$ and a $1/r^2$ decrease for $r > R$ is the correct one.

(A) The electric field inside a uniformly charged insulating sphere at distance $r < R$ is given by $E = \frac{\rho r}{3 \epsilon_0}$. Thus,Statement-$2$ is true.
The electric potential $V(r)$ inside a uniformly charged sphere is given by $V(r) = \frac{\rho}{6 \epsilon_0} (3R^2 - r^2)$.
At the centre $(r = 0)$,$V_{centre} = \frac{3 \rho R^2}{6 \epsilon_0} = \frac{\rho R^2}{2 \epsilon_0}$.
At the surface $(r = R)$,$V_{surface} = \frac{\rho}{6 \epsilon_0} (3R^2 - R^2) = \frac{2 \rho R^2}{6 \epsilon_0} = \frac{\rho R^2}{3 \epsilon_0}$.
The change in potential energy $\Delta U = q(V_{surface} - V_{centre}) = q \left( \frac{\rho R^2}{3 \epsilon_0} - \frac{\rho R^2}{2 \epsilon_0} \right) = q \left( \frac{2 \rho R^2 - 3 \rho R^2}{6 \epsilon_0} \right) = -\frac{q \rho R^2}{6 \epsilon_0}$.
Since the magnitude of change is $\frac{q \rho R^2}{6 \epsilon_0}$,Statement-$1$ is true. Statement-$2$ provides the formula for the electric field,which is used to derive the potential difference,making it the correct explanation.

(C) Consider a Gaussian surface as a sphere of radius $r$ such that $a < r < b$.
According to Gauss's law,$\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}$.
Here,$q_{enclosed} = Q + \int_{a}^{r} \rho(r') \cdot 4\pi r'^2 dr'$.
Given $\rho(r') = \frac{A}{r'}$,we have $q_{enclosed} = Q + \int_{a}^{r} \frac{A}{r'} \cdot 4\pi r'^2 dr' = Q + 4\pi A \int_{a}^{r} r' dr' = Q + 4\pi A \left[ \frac{r'^2}{2} \right]_{a}^{r} = Q + 2\pi A (r^2 - a^2)$.
Applying Gauss's law: $E \cdot 4\pi r^2 = \frac{Q + 2\pi A (r^2 - a^2)}{\epsilon_0}$.
$E = \frac{1}{4\pi \epsilon_0 r^2} [Q - 2\pi A a^2 + 2\pi A r^2] = \frac{1}{4\pi \epsilon_0} [\frac{Q - 2\pi A a^2}{r^2} + 2\pi A]$.
For the electric field $E$ to be constant (independent of $r$),the coefficient of $\frac{1}{r^2}$ must be zero.
Therefore,$Q - 2\pi A a^2 = 0$.
$A = \frac{Q}{2\pi a^2}$.

(A) According to Gauss's Law,the total electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$.
For a sphere of radius $R$ centered at the origin,the electric field is radially outward and its magnitude is $E = \alpha R$ at the surface.
The area of the sphere is $A = 4\pi R^2$.
The flux through the sphere is $\Phi_E = E \times A = (\alpha R) \times (4\pi R^2) = 4\pi \alpha R^3$.
Using Gauss's Law,$Q_{enclosed} = \varepsilon_0 \Phi_E$.
Therefore,$Q = \varepsilon_0 (4\pi \alpha R^3) = 4\pi \varepsilon_0 \alpha R^3$.

(C) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is given by $\phi = \oint \vec{E} \cdot d\vec{s} = \frac{Q_{in}}{\varepsilon_0}$.
For a cylinder with its axis along the $y$-axis, the electric field $\vec{E} = E_0 y \hat{j}$ is parallel to the curved surface and perpendicular to the circular end faces.
However, the flux through the curved surface is zero because $\vec{E} \cdot d\vec{s} = 0$ (as $d\vec{s}$ is radial).
At the bottom face $(y = 0)$, $\vec{E} = 0$, so flux is $0$.
At the top face $(y = l)$, $\vec{E} = E_0 l \hat{j}$ and the area vector $d\vec{s} = dA \hat{j}$.
The flux through the top face is $\phi = \int E_0 l dA = E_0 l \int dA = E_0 l (\pi r^2)$.
Thus, the total flux $\phi = E_0 l \pi r^2$.
Using Gauss's Law, $Q_{in} = \varepsilon_0 \phi = \varepsilon_0 E_0 \pi r^2 l$.

(B) The electric field due to an infinite line charge at a distance $r$ is $E = \frac{\lambda}{2\pi \varepsilon_0 r}$.
Since the electric field is conservative,the work done in moving a charge $q$ depends only on the initial and final positions.
Point $A$ is at a distance $r_A = a$ from the $x$-axis.
Point $C$ is at a distance $r_C = \sqrt{a^2 + a^2} = a\sqrt{2}$ from the $x$-axis.
The work done $W_{CA}$ is given by $W = q(V_A - V_C) = -q \int_{r_C}^{r_A} E dr$.
$W = -q \int_{a\sqrt{2}}^{a} \frac{\lambda}{2\pi \varepsilon_0 r} dr = -\frac{q\lambda}{2\pi \varepsilon_0} [\ln r]_{a\sqrt{2}}^{a}$.
$W = -\frac{q\lambda}{2\pi \varepsilon_0} (\ln a - \ln(a\sqrt{2})) = -\frac{q\lambda}{2\pi \varepsilon_0} \ln(\frac{a}{a\sqrt{2}})$.
$W = -\frac{q\lambda}{2\pi \varepsilon_0} \ln(\frac{1}{\sqrt{2}}) = \frac{q\lambda}{2\pi \varepsilon_0} \ln(\sqrt{2}) = \frac{q\lambda}{2\pi \varepsilon_0} \ln(2^{1/2}) = \frac{q\lambda}{4\pi \varepsilon_0} \ln 2$.

(A) For $r < a$,the electric field inside a conducting shell is zero,so $E = 0$.
For $a < r < b$,the electric field is due to the inner shell only. Using Gauss's Law,$E = \frac{kQ}{r^2}$.
For $r > b$,the total charge enclosed is $Q + (-2Q) = -Q$. Thus,the electric field is $E = \frac{k(-Q)}{r^2} = -\frac{kQ}{r^2}$.
Comparing these results with the given options,the graph showing $E = 0$ for $r < a$,a positive $1/r^2$ variation for $a < r < b$,and a negative $1/r^2$ variation for $r > b$ is represented by the first graph.

(B) For a point $A$ inside the sphere $(x < R)$,using Gauss's Law: $E_A \cdot 4\pi x^2 = \frac{q_{enclosed}}{\epsilon_0} = \frac{1}{\epsilon_0} \int_0^x \rho(r) \cdot 4\pi r^2 dr$.
Substituting $\rho = \rho_0 r^2$: $E_A \cdot 4\pi x^2 = \frac{4\pi \rho_0}{\epsilon_0} \int_0^x r^4 dr = \frac{4\pi \rho_0 x^5}{5\epsilon_0}$.
Thus,$E_A = \frac{\rho_0 x^3}{5\epsilon_0}$.
For a point $B$ outside the sphere $(y > R)$,the sphere acts as a point charge at the centre: $E_B = \frac{Q_{total}}{4\pi \epsilon_0 y^2}$.
$Q_{total} = \int_0^R \rho_0 r^2 \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_0^R r^4 dr = \frac{4\pi \rho_0 R^5}{5}$.
So,$E_B = \frac{4\pi \rho_0 R^5}{5 \cdot 4\pi \epsilon_0 y^2} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$.
Given $E_A = E_B$,we have $\frac{\rho_0 x^3}{5\epsilon_0} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$.
Simplifying,we get $x^3 y^2 = R^5$.

(B) According to Gauss's law, the electric field inside the conducting material of the cylinder must be zero.
Since the wire has a charge per unit length of $\lambda$, an induced charge of $-\lambda$ per unit length must appear on the inner surface of the cylinder.
The surface charge density on the inner surface is $\sigma_{in} = \frac{-\lambda}{2\pi R}$.
Since the net charge per unit length of the cylinder is $2\lambda$, the charge on the outer surface per unit length is $2\lambda - (-\lambda) = 3\lambda$.
The surface charge density on the outer surface is $\sigma_{out} = \frac{3\lambda}{2\pi R}$.
For a Gaussian surface of radius $r > R$, the total enclosed charge per unit length is $\lambda + 3\lambda = 4\lambda$.
Applying Gauss's law: $E(2\pi r) = \frac{4\lambda}{\epsilon_0} \implies E = \frac{2\lambda}{\pi \epsilon_0 r}$.
Comparing with the options, statement $B$ is correct.

(D) The sheet produces a uniform electric field $E = \frac{\sigma}{2\epsilon_0}$ directed away from the sheet.
The force on the left half $(AC)$ is $F = qE = (\lambda \cdot \frac{l}{2}) \cdot \frac{\sigma}{2\epsilon_0} = \frac{\lambda l \sigma}{4\epsilon_0}$ acting to the left.
The force on the right half $(CB)$ is $F = qE = (\lambda \cdot \frac{l}{2}) \cdot \frac{\sigma}{2\epsilon_0} = \frac{\lambda l \sigma}{4\epsilon_0}$ acting to the right.
When the rod is rotated by a small angle $\theta$,the torque about the hinge $C$ is $\tau = 2 \cdot F \cdot (\frac{l}{4} \cos \theta) \approx 2 \cdot F \cdot \frac{l}{4} = \frac{Fl}{2} = \frac{\lambda l^2 \sigma}{8\epsilon_0}$.
Since $\tau = I\alpha$,where $I = \frac{ml^2}{12}$ is the moment of inertia about the center $C$,we have $\frac{ml^2}{12} \alpha = \frac{\lambda l^2 \sigma}{8\epsilon_0} \theta$.
$\alpha = \frac{12 \lambda \sigma}{8 m \epsilon_0} \theta = \frac{3 \lambda \sigma}{2 m \epsilon_0} \theta$.
Comparing with $\alpha = \omega^2 \theta$,we get $\omega^2 = \frac{3 \lambda \sigma}{2 m \epsilon_0}$.
The time period is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2m\epsilon_0}{3\lambda\sigma}}$.

(B) According to Gauss's Law,the total electric flux through a closed surface is $\phi_{total} = \frac{q_{enclosed}}{\varepsilon_0}$.
In this problem,the charges $3q$ and $5q$ are placed on the planar surface of the hemisphere. Since they are on the boundary,we can imagine a complete sphere by placing an identical hemisphere on top of the given one.
The total charge enclosed by this complete sphere is $Q_{enclosed} = 3q + 5q = 8q$.
By symmetry,the flux through the original hemisphere is half of the total flux through the complete sphere.
Therefore,the electric flux over the curved surface is $\phi = \frac{1}{2} \times \frac{Q_{enclosed}}{\varepsilon_0} = \frac{1}{2} \times \frac{8q}{\varepsilon_0} = \frac{4q}{\varepsilon_0}$.

(B) $1$. According to the properties of a conductor in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.
$2$. Consider a Gaussian surface inside the conducting material of the shell. By Gauss's Law,the total charge enclosed by this surface must be zero.
$3$. The total charge enclosed is the sum of the point charge $+Q$ at the center and the induced charge $q_{in}$ on the inner surface of the shell. Thus,$+Q + q_{in} = 0$,which implies $q_{in} = -Q$.
$4$. The shell has a total charge of $-q$. Let $q_{out}$ be the charge on the outer surface. Since the total charge is the sum of the charges on the inner and outer surfaces,we have $q_{in} + q_{out} = -q$.
$5$. Substituting $q_{in} = -Q$,we get $-Q + q_{out} = -q$,which gives $q_{out} = -q + Q$.
$6$. Therefore,the charge on the inner surface is $-Q$ and the charge on the outer surface is $(-q + Q)$.

(A) The electric field just outside a charged conductor is given by $E = \sigma / \epsilon_0$,where $\sigma$ is the surface charge density.
This field is the result of the superposition of the field due to the small patch where the hole is made $(E_{patch})$ and the field due to the rest of the shell $(E_{rest})$.
At a point just outside the surface,$E_{patch} + E_{rest} = E = \sigma / \epsilon_0$.
Since the field due to a small patch is $E_{patch} = \sigma / (2\epsilon_0)$,we have $E_{rest} = E - E_{patch} = \sigma / (2\epsilon_0) = E/2$.
At a point just inside the surface,the fields are $E_{patch} - E_{rest} = \sigma / (2\epsilon_0) - \sigma / (2\epsilon_0) = 0$.
When a hole is made,the field inside the hole is due to the rest of the shell $(E_{rest})$,which is $E/2$.

(C) For a large conducting plate (copper),the charge $Q$ distributes on both sides of the plate. The surface charge density on each side is $\sigma = \frac{Q}{2A}$. The electric field near the surface is $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{2A\epsilon_0} = 10 \ V/m$.
For a non-conducting plate (plastic),the charge $Q$ is fixed on the surface where it is placed. If the charge is distributed over the entire surface area $A$,the surface charge density is $\sigma' = \frac{Q}{A}$. The electric field due to a non-conducting sheet is $E' = \frac{\sigma'}{2\epsilon_0} = \frac{Q}{2A\epsilon_0}$.
Comparing the two expressions,we see that $E' = E = 10 \ V/m$.

(B) Consider a solid sphere of radius $R$ with uniform charge density $\rho$. The electric field at a point $\vec{r}$ inside the sphere is given by $\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$.
When a smaller spherical cavity is created,the electric field in the cavity is the superposition of the field due to the original large sphere and the field due to a smaller sphere of opposite charge density $-\rho$ that fills the cavity.
Let $\vec{r}_1$ be the position vector from the center of the large sphere and $\vec{r}_2$ be the position vector from the center of the cavity.
The electric field in the cavity is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\rho \vec{r}_1}{3\epsilon_0} + \frac{-\rho \vec{r}_2}{3\epsilon_0} = \frac{\rho}{3\epsilon_0} (\vec{r}_1 - \vec{r}_2)$.
Since $(\vec{r}_1 - \vec{r}_2) = \vec{d}$,where $\vec{d}$ is the constant vector connecting the center of the large sphere to the center of the cavity,the electric field $\vec{E} = \frac{\rho \vec{d}}{3\epsilon_0}$ is constant (uniform) and non-zero throughout the cavity.

(D) The total charge $q$ inside the ball is found by integrating the charge density over the volume of the ball:
$q = \int_0^R \rho(r) \cdot 4\pi r^2 dr$
$q = \int_0^R \rho_0 \left( 1 - \frac{r}{R} \right) 4\pi r^2 dr$
$q = 4\pi \rho_0 \int_0^R \left( r^2 - \frac{r^3}{R} \right) dr$
$q = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_0^R$
$q = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) = 4\pi \rho_0 \left( \frac{R^3}{12} \right) = \frac{\pi \rho_0 R^3}{3}$
Using Gauss's Law for a point outside the ball at distance $r$ $(r > R)$:
$E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}$
$E \cdot 4\pi r^2 = \frac{\pi \rho_0 R^3}{3 \varepsilon_0}$
$E = \frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$

(C) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.
Given that the difference in radii $\Delta r = r_{i+1} - r_i$ is constant for a constant potential difference $\Delta V$,it implies that the electric field $E = -\frac{\Delta V}{\Delta r}$ is constant.
For a spherical charge distribution,by Gauss's Law,the electric field at a distance $r$ is $E = \frac{q_{enclosed}}{4\pi\epsilon_0 r^2}$.
Since $E$ is constant,we have $q_{enclosed} \propto r^2$.
We know that $q_{enclosed} = \int_0^r \rho(r) 4\pi r^2 dr$.
Since $q_{enclosed} \propto r^2$,differentiating both sides with respect to $r$ gives $\frac{dq}{dr} \propto 2r$.
Thus,$\rho(r) 4\pi r^2 \propto r$,which implies $\rho(r) \propto \frac{1}{r}$.

(A) The electric field $E$ on the axis of a disc of radius $R$ at a distance $h$ from its center is given by $E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{R^2 + h^2}} \right)$.
For the disc with a hole,we can consider it as a large disc of radius $R_1 = 2a$ minus a smaller disc of radius $R_2 = a$.
The electric field $E_1$ due to the large disc is $E_1 = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{(2a)^2 + h^2}} \right) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{2a \sqrt{1 + (h/2a)^2}} \right)$.
Since $h << a$,we use the binomial approximation $(1+x)^n \approx 1+nx$: $E_1 \approx \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{2a} \right)$.
The electric field $E_2$ due to the smaller disc is $E_2 = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{a^2 + h^2}} \right) \approx \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{a} \right)$.
The net electric field $E = E_1 - E_2 = \frac{\sigma}{2\epsilon_0} \left[ (1 - \frac{h}{2a}) - (1 - \frac{h}{a}) \right] = \frac{\sigma}{2\epsilon_0} \left( \frac{h}{a} - \frac{h}{2a} \right) = \frac{\sigma}{2\epsilon_0} \left( \frac{h}{2a} \right) = \frac{\sigma h}{4a\epsilon_0}$.
Comparing this with $Ch$,we get $C = \frac{\sigma}{4a\epsilon_0}$.

(C) Given:
Electric field $E = 150\, N/C$ (directed inward,so $E = -150\, N/C$)
Radius of Earth $R_E = 6.37 \times 10^6\, m$
Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}\, C^2/(N \cdot m^2)$
According to Gauss's law,the electric flux $\phi$ through the surface of the Earth is $\phi = \frac{q}{\epsilon_0} = E \cdot A$,where $A = 4\pi R_E^2$ is the surface area of the Earth.
Thus,$q = \epsilon_0 \cdot E \cdot 4\pi R_E^2$
Substituting the values:
$q = (8.85 \times 10^{-12}) \times (-150) \times 4 \times 3.14159 \times (6.37 \times 10^6)^2$
$q \approx -6.80 \times 10^5\, C$
Converting to $kC$ $(1\, kC = 10^3\, C)$:
$q \approx -680\, kC$
Since the electric field is directed inward,the charge is negative.

(B) According to Gauss's Law,the electric field $E$ at a distance $r$ from the centre of a spherically symmetric charge distribution is given by $E \cdot (4\pi r^2) = \frac{q_{enclosed}}{\varepsilon_0}$.
To find the enclosed charge $q$ within a sphere of radius $r < R$,we integrate the charge density over the volume:
$q = \int_0^r \rho(x) \cdot 4\pi x^2 dx$
Substituting $\rho(x) = \rho_0 \left( 1 - \frac{x}{R} \right)$:
$q = 4\pi \rho_0 \int_0^r \left( x^2 - \frac{x^3}{R} \right) dx$
$q = 4\pi \rho_0 \left[ \frac{x^3}{3} - \frac{x^4}{4R} \right]_0^r = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$
Now,using Gauss's Law:
$E \cdot 4\pi r^2 = \frac{4\pi \rho_0}{\varepsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$
Dividing both sides by $4\pi r^2$:
$E = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right)$

(A) The electric field at the centre of a disc is $E = \frac{\sigma}{2\epsilon_0}$.
The electric field at a distance $x$ along the axis of a disc is given by $E' = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$.
Given $x = R$,we substitute this into the formula:
$E' = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{R^2 + R^2}}\right) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{2}R}\right) = E \left(1 - \frac{1}{\sqrt{2}}\right)$.
Since $\frac{1}{\sqrt{2}} \approx 0.707$,we have $E' = E(1 - 0.707) = 0.293E$.
The reduction in the electric field is $E - E' = E - 0.293E = 0.707E$.
The percentage reduction is $\frac{0.707E}{E} \times 100\% = 70.7\%$.

(C) The electric field $E$ is given by the negative gradient of the potential: $E = -\frac{dV}{dx} = -\frac{d}{dx}(4x^2) = -8x \text{ V/m}$.
According to Gauss's Law,the total flux $\phi$ through a closed surface is $\phi = \oint E \cdot dA = \frac{q_{\text{enclosed}}}{\varepsilon_0}$.
For a cube of side $L = 1 \text{ m}$ centered at the origin,the faces are at $x = 0.5 \text{ m}$ and $x = -0.5 \text{ m}$.
The electric field at $x = 0.5 \text{ m}$ is $E_1 = -8(0.5) = -4 \text{ V/m}$ (pointing towards the origin).
The electric field at $x = -0.5 \text{ m}$ is $E_2 = -8(-0.5) = 4 \text{ V/m}$ (pointing away from the origin).
The flux through the face at $x = 0.5 \text{ m}$ is $\phi_1 = E_1 \cdot A = -4 \times (1^2) = -4 \text{ V} \cdot \text{m}^2$.
The flux through the face at $x = -0.5 \text{ m}$ is $\phi_2 = E_2 \cdot A = 4 \times (1^2) = 4 \text{ V} \cdot \text{m}^2$.
The flux through the other four faces is zero because the electric field is parallel to these faces.
The net flux is $\phi_{\text{net}} = \phi_1 + \phi_2 = -4 + 4 = 0$.
Since $\phi_{\text{net}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$,we have $q_{\text{enclosed}} = 0$.

(C) The electric field $E$ at a distance $r$ from a line charge with linear charge density $\lambda$ is given by $E = \frac{\lambda}{2\pi\varepsilon_0 r}$.
The potential difference $\Delta V$ between distance $r_0$ and $r$ is $\Delta V = -\int_{r_0}^{r} E \, dr = -\int_{r_0}^{r} \frac{\lambda}{2\pi\varepsilon_0 r} \, dr = -\frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r}{r_0} \right) = \frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r_0}{r} \right)$.
Using the work-energy theorem,the change in kinetic energy equals the work done by the electric field: $\frac{1}{2}mv^2 = q(V_i - V_f) = q \Delta V$.
Substituting the potential difference: $\frac{1}{2}mv^2 = q \left( \frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r}{r_0} \right) \right)$.
Since $m, q, \lambda, \pi, \varepsilon_0$ are constants,we have $v^2 \propto \ln \left( \frac{r}{r_0} \right)$,which implies $v \propto \sqrt{\ln \left( \frac{r}{r_0} \right)}$.

(C) First,we find the constant $k$ in terms of $Q$ and $R$ by integrating the charge density over the volume of the sphere:
$2Q = \int_{0}^{R} \rho(r) 4\pi r^2 dr = \int_{0}^{R} (kr) 4\pi r^2 dr = 4\pi k \int_{0}^{R} r^3 dr = 4\pi k \frac{R^4}{4} = \pi k R^4$.
Thus,$k = \frac{2Q}{\pi R^4}$.
Next,we find the electric field $E$ at distance $a$ from the centre using Gauss's Law:
$E(4\pi a^2) = \frac{q_{enc}}{\varepsilon_0} = \frac{1}{\varepsilon_0} \int_{0}^{a} (kr) 4\pi r^2 dr = \frac{4\pi k}{\varepsilon_0} \frac{a^4}{4} = \frac{\pi k a^4}{\varepsilon_0}$.
$E = \frac{k a^2}{4\varepsilon_0} = \frac{(2Q/\pi R^4) a^2}{4\varepsilon_0} = \frac{Q a^2}{2\pi \varepsilon_0 R^4}$.
For charge $A$ (or $B$) to experience no force,the force due to the sphere must be balanced by the force due to the other charge:
$|qE| = \frac{1}{4\pi \varepsilon_0} \frac{|Q||-Q|}{(2a)^2} \implies Q \left( \frac{Q a^2}{2\pi \varepsilon_0 R^4} \right) = \frac{Q^2}{16\pi \varepsilon_0 a^2}$.
Simplifying: $\frac{a^2}{2 R^4} = \frac{1}{16 a^2} \implies a^4 = \frac{R^4}{8} \implies a = \frac{R}{8^{1/4}} = 8^{-1/4}R$.

(C) According to Gauss's Law,for a non-conducting sphere with uniform charge density $\rho$,the electric flux through a Gaussian surface of radius $r$ (where $r < R$) is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.
Since the sphere is non-conducting and the charge is distributed throughout the volume,the enclosed charge $q_{enclosed} = \rho \cdot V = \rho \cdot (\frac{4}{3}\pi r^3)$.
The surface area of the Gaussian sphere is $4\pi r^2$.
Thus,$E \cdot (4\pi r^2) = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$.
Solving for $E$,we get $E = \frac{\rho r}{3\varepsilon_0}$.

(C) According to Gauss's Law,the electric field $E$ at a distance $r_1$ from the center is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.
For a spherical surface of radius $r_1$,this becomes $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \rho(r) (4\pi r^2) dr$.
Substituting $\rho(r) = \frac{Q}{\pi R^4} r$,we get $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \left( \frac{Q}{\pi R^4} r \right) (4\pi r^2) dr$.
$E(4\pi r_1^2) = \frac{4\pi Q}{\pi R^4 \varepsilon_0} \int_0^{r_1} r^3 dr$.
$E(4\pi r_1^2) = \frac{4 Q}{R^4 \varepsilon_0} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4 \varepsilon_0}$.
Solving for $E$,we get $E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

(D) To find the electric field at a distance $r_1$ from the center,we first calculate the charge $q$ enclosed within a sphere of radius $r_1$.
The charge element $dq$ in a spherical shell of radius $r$ and thickness $dr$ is given by $dq = \rho(r) \cdot 4\pi r^2 dr$.
Substituting $\rho(r) = \frac{Q}{\pi R^4} r$,we get:
$dq = \left( \frac{Q}{\pi R^4} r \right) \cdot 4\pi r^2 dr = \frac{4Q}{R^4} r^3 dr$.
The total charge $q$ enclosed within radius $r_1$ is:
$q = \int_0^{r_1} \frac{4Q}{R^4} r^3 dr = \frac{4Q}{R^4} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4}$.
Using Gauss's Law for a spherical surface of radius $r_1$:
$E \cdot 4\pi r_1^2 = \frac{q}{\varepsilon_0}$.
Substituting the value of $q$:
$E \cdot 4\pi r_1^2 = \frac{Q r_1^4}{\varepsilon_0 R^4}$.
Solving for $E$:
$E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

(C) When sphere $A$ is given a charge $Q$,it induces a charge $-Q$ on the inner surface of the earthed sphere $B$.
Since sphere $B$ is earthed,its potential becomes zero.
The electric field inside a conductor is zero,but in the region between the two spheres (where $r_A < r < r_B$),the electric field is due to the charge $Q$ on sphere $A$.
Using Gauss's Law,the electric field $E$ at a distance $r$ from the center is $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$,which is non-zero.
Outside the outer sphere $B$,the net charge is $Q + (-Q) = 0$,so the electric field outside $B$ is zero.
Thus,the field between $A$ and $B$ is not zero,and the field outside $B$ is zero.

(C) For a uniformly charged non-conducting sphere of radius $R$ and total charge $Q$:
$1$. Inside the sphere $(r < R)$, the electric field is given by $E_{in} = \frac{kQr}{R^3}$, which means $E_{in} \propto r$. This is a linear relationship starting from the origin.
$2$. Outside the sphere $(r \geq R)$, the electric field is given by $E_{out} = \frac{kQ}{r^2}$, which means $E_{out} \propto \frac{1}{r^2}$. This is an inverse square relationship.
$3$. At the surface $(r = R)$, the electric field is maximum, $E_{max} = \frac{kQ}{R^2}$.
Combining these, the graph starts linearly from the origin up to $r = R$ and then follows an inverse square curve for $r > R$. This corresponds to graph $C$.

(A) For a uniformly charged non-conducting sphere of radius $R$ and total charge $Q$,the volume charge density is $\rho = \frac{Q}{\frac{4}{3}\pi R^3}$.
Using Gauss's Law for a Gaussian surface of radius $r$ $(r < R)$:
$\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$
$E(4\pi r^2) = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\epsilon_0}$
Substituting $\rho = \frac{Q}{\frac{4}{3}\pi R^3}$:
$E(4\pi r^2) = \frac{Q}{\frac{4}{3}\pi R^3} \cdot \frac{\frac{4}{3}\pi r^3}{\epsilon_0} = \frac{Q r^3}{\epsilon_0 R^3}$
$E = \frac{Q r}{4\pi \epsilon_0 R^3}$
Since $K = \frac{1}{4\pi \epsilon_0}$,we get $E = \frac{KQr}{R^3}$.

(D) The electric field $E$ at a perpendicular distance $r$ from an infinitely long uniformly charged wire is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
For point $A(3, 4, -6)$,the perpendicular distance $r_A$ from the $z$-axis is $r_A = \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
For point $B(3, 4, 0)$,the perpendicular distance $r_B$ from the $z$-axis is $r_B = \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
Since $r_A = r_B = 5$,the electric fields at points $A$ and $B$ are equal.
Therefore,the ratio of the electric field at $A$ and $B$ is $E_A : E_B = 1 : 1$.

(B) For an infinitely long uniformly charged wire with linear charge density $\lambda$,we use Gauss's Law to find the electric field at a distance $r$.
Consider a cylindrical Gaussian surface of radius $r$ and length $l$ coaxial with the wire.
The total electric flux $\Phi_E$ through the Gaussian surface is given by $\Phi_E = E \times (2\pi rl)$.
According to Gauss's Law,$\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$.
Since the charge enclosed is $q_{enclosed} = \lambda l$,we have $E \times (2\pi rl) = \frac{\lambda l}{\epsilon_0}$.
Solving for $E$,we get $E = \frac{\lambda}{2\pi \epsilon_0 r}$.
Therefore,the electric field $E$ is inversely proportional to the distance $r$,i.e.,$E \propto \frac{1}{r}$.

(C) The electric field $E$ between two large parallel thin metal sheets with equal and opposite surface charge densities $\sigma$ is given by the formula:
$E = \frac{\sigma}{\epsilon_0}$
Given:
$\sigma = 26.4 \times 10^{-12} \, C/m^2$
$\epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)$
Substituting the values:
$E = \frac{26.4 \times 10^{-12}}{8.85 \times 10^{-12}}$
$E = \frac{26.4}{8.85} \approx 2.983 \approx 3 \, N/C$
Therefore,the electric field between the sheets is $3 \, N/C$.

(C) According to Gauss's Law,the electric flux through a closed Gaussian surface is given by $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
For a point at a distance $r$ such that $r_1 < r < r_2$,we consider a spherical Gaussian surface of radius $r$ concentric with the shells.
The only charge enclosed by this Gaussian surface is the charge $Q_1$ on the inner shell.
Therefore,$Q_{\text{enclosed}} = Q_1$.
Applying Gauss's Law: $E(4\pi r^2) = \frac{Q_1}{\epsilon_0}$.
Thus,the electric field is $E = \frac{Q_1}{4\pi \epsilon_0 r^2}$.

(A) To find the electric field at point $P$,we use the principle of superposition. We consider the cylinder with the cavity as the sum of a complete solid cylinder with charge density $\rho$ and a sphere with charge density $-\rho$.
$1$. Electric field due to the infinite solid cylinder of radius $R$ at distance $r = 2R$ from the axis:
Using Gauss's Law,$E_{cyl} \cdot (2\pi r L) = \frac{q_{enclosed}}{\varepsilon_0} = \frac{\rho \cdot \pi R^2 L}{\varepsilon_0}$.
$E_{cyl} = \frac{\rho R^2}{2 \varepsilon_0 r} = \frac{\rho R^2}{2 \varepsilon_0 (2R)} = \frac{\rho R}{4 \varepsilon_0}$.
$2$. Electric field due to the spherical cavity of radius $a = R/2$ at distance $r = 2R$ from its center:
Using the formula for the electric field outside a uniformly charged sphere,$E_{sph} = \frac{kQ}{r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{\rho \cdot (4/3)\pi a^3}{r^2}$.
Substituting $a = R/2$ and $r = 2R$:
$E_{sph} = \frac{\rho \cdot (4/3)\pi (R/2)^3}{4\pi \varepsilon_0 (2R)^2} = \frac{\rho \cdot (4/3) (R^3/8)}{4 \varepsilon_0 (4R^2)} = \frac{\rho R^3 / 6}{16 \varepsilon_0 R^2} = \frac{\rho R}{96 \varepsilon_0}$.
$3$. Net electric field at $P$:
Since the sphere has charge density $-\rho$,the field is in the opposite direction to the cylinder's field.
$E_{net} = E_{cyl} - E_{sph} = \frac{\rho R}{4 \varepsilon_0} - \frac{\rho R}{96 \varepsilon_0} = \frac{24\rho R - \rho R}{96 \varepsilon_0} = \frac{23\rho R}{96 \varepsilon_0}$.
Comparing this with the given expression $\frac{23\rho R}{16K\varepsilon_0}$:
$16K = 96 \implies K = 6$.