Conductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor Questions in English

(A) When two plates of a capacitor have charges $Q_1$ and $Q_2$,the charge $Q$ that appears on the inner facing surfaces is given by $Q = \frac{Q_1 - Q_2}{2}$.
Here,$Q_1 = 15\,\mu C$ and $Q_2 = 5\,\mu C$.
So,the charge on the inner surfaces is $Q = \frac{15\,\mu C - 5\,\mu C}{2} = \frac{10\,\mu C}{2} = 5\,\mu C$.
The potential difference $V$ across the capacitor is given by $V = \frac{Q}{C}$.
Given $C = 10\,\mu F = 10 \times 10^{-6}\,F$ and $Q = 5\,\mu C = 5 \times 10^{-6}\,C$.
Therefore,$V = \frac{5 \times 10^{-6}}{10 \times 10^{-6}} = 0.5\,V$.

(C) Initial charge on sphere $A$ $(q_A)$ $= 7\,\mu C$.
Initial charge on sphere $B$ $(q_B)$ $= 1\,\mu C$.
Since the spheres are identical,when they are connected by a wire,the total charge is redistributed equally between them.
Total charge $Q = q_A + q_B = 7\,\mu C + 1\,\mu C = 8\,\mu C$.
Final charge on each sphere after connection $q' = \frac{Q}{2} = \frac{8\,\mu C}{2} = 4\,\mu C$.
The charge flow from $A$ to $B$ is the difference between the initial and final charge on sphere $A$.
Charge flow $= q_A - q' = 7\,\mu C - 4\,\mu C = 3\,\mu C$.

(C) When a conductor is placed in an external uniform electric field,the free electrons within the conductor redistribute themselves until the internal electric field becomes zero.
For any conductor,the induced surface charge density $\sigma$ is given by $\sigma = \epsilon_0 E_n$,where $E_n$ is the component of the external electric field normal to the surface.
Since both conductors have the same shape and size and are placed in the same uniform electric field,the external field $E$ at the surface is identical for both.
Therefore,the total induced charge $q = \int \sigma dA$ will be the same for both the copper and the aluminium conductors,regardless of their material conductivity,as long as they are both conductors.

(C) According to the principle of electrostatic shielding,the electric field inside a cavity of a conductor is zero,regardless of any external charges or electric fields present outside the conductor.
When a charge $Q$ is brought outside the metal block,the charges on the outer surface of the metal redistribute themselves to ensure that the electric field inside the metal remains zero.
However,the electric field inside the cavity is determined solely by the charge $q$ placed inside it and the induced charges on the inner surface of the cavity.
The external charge $Q$ does not affect the electric field inside the cavity because the metal block acts as an electrostatic shield.
Therefore,the electric force experienced by the point charge $q$ remains unaffected by the external charge $Q$ and is zero if $q$ is at the center of a spherical cavity or depends only on the local configuration of induced charges on the inner wall,but the net force due to external charge $Q$ is zero.

(D) The potential of a charged sphere of radius $r$ with charge $q$ is given by $V = \frac{kq}{r}$.
For the first sphere,the potential is $V_1 = \frac{kQ}{R}$.
For the second sphere,the potential is $V_2 = \frac{k(2Q)}{2R} = \frac{kQ}{R}$.
Since the potentials of both spheres are equal $(V_1 = V_2)$,there is no potential difference between them.
Therefore,when the key $k$ is pressed,no charge will flow between the two spheres.

(D) Let the charge on the inner shell after closing the key be $q_1'$.
Since the inner shell is connected to the ground, its potential becomes zero.
The potential of the inner shell is given by the sum of potentials due to its own charge and the charge on the outer shell:
$V_1 = \frac{k q_1'}{R_1} + \frac{k q_2}{R_2} = 0$
From this, we get:
$\frac{q_1'}{R_1} = -\frac{q_2}{R_2} \implies q_1' = -q_2 \frac{R_1}{R_2}$
The charge that flows through the key is the change in charge on the inner shell:
$\Delta q = q_{final} - q_{initial} = q_1' - q_1$
$\Delta q = -q_2 \frac{R_1}{R_2} - q_1 = -(q_1 + q_2 \frac{R_1}{R_2})$
Thus, the magnitude of charge flowing to the ground is $(q_1 + q_2 \frac{R_1}{R_2})$.

(A) According to the principle of electrostatic shielding,the electric field inside a closed metallic conductor is always zero,regardless of any external electric field.
This phenomenon occurs because the free electrons in the metal redistribute themselves on the surface to cancel out the effect of the external field.
Since the electric field inside the hollow shell is zero,it effectively blocks external electric fields from entering the interior region.
Therefore,the Assertion is correct,and the Reason provides the correct physical explanation for why such a shield works.

(A) In a conductor,the net electric field inside the material is zero because the free electrons redistribute themselves to cancel any external electric field.
Due to this redistribution,the net charge inside the volume of the conductor is zero,and all excess charges reside on the outer surface of the conductor.
According to Gauss's Law,if there is no charge enclosed within a cavity inside a conductor,the electric field inside that cavity must also be zero.
Thus,the reason correctly explains why the electric field is zero in the cavity,as the charges reside only on the surface,leaving the interior (including the cavity) field-free.

(D) For any closed surface enclosing a net charge of $0$, the total electric flux through the surface is $0$ according to Gauss's Law $(\Phi = \oint E \cdot dA = q_{enclosed} / \epsilon_0 = 0)$.
This means the number of electric field lines entering the surface must exactly equal the number of lines leaving the surface.
Therefore, the Assertion is incorrect because the number of lines terminating on the sphere cannot be more than those emerging from it.
The Reason is correct because, due to electrostatic induction, the side of the conductor closer to the positive point charge develops a negative induced charge, which has the highest surface charge density in magnitude.
Since the Assertion is incorrect and the Reason is correct, the correct choice is $D$ (based on standard logic for this specific question type).

(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.
Total charge $Q_{total} = Q_1 + Q_2 = 4 \pi R^2 \sigma + 16 \pi R^2 \sigma = 20 \pi R^2 \sigma$.
When the spheres are brought into contact,the total charge is redistributed such that they reach the same potential $V = \frac{k Q_1'}{R} = \frac{k Q_2'}{2R}$.
This implies $Q_2' = 2 Q_1'$.
Since $Q_1' + Q_2' = 20 \pi R^2 \sigma$,we have $Q_1' + 2 Q_1' = 20 \pi R^2 \sigma$,which gives $3 Q_1' = 20 \pi R^2 \sigma$,so $Q_1' = \frac{20}{3} \pi R^2 \sigma$.
Then $Q_2' = 2 \times \frac{20}{3} \pi R^2 \sigma = \frac{40}{3} \pi R^2 \sigma$.
The new surface charge densities are $\sigma_1' = \frac{Q_1'}{4 \pi R^2} = \frac{20/3 \pi R^2 \sigma}{4 \pi R^2} = \frac{5}{3} \sigma$.
And $\sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} = \frac{40/3 \pi R^2 \sigma}{16 \pi R^2} = \frac{40}{48} \sigma = \frac{5}{6} \sigma$.

(N/A) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity $E$ inside the charged conductor is zero. Let $q_{in}$ be the charge inside the conductor and $\epsilon_{0}$ be the permittivity of free space. According to Gauss's law, the electric flux $\phi$ is given by:
$\phi = \oint E \cdot ds = \frac{q_{in}}{\epsilon_{0}}$
Since $E = 0$ inside the conductor, we have $\frac{q_{in}}{\epsilon_{0}} = 0$, which implies $q_{in} = 0$.
Therefore, the net charge inside the material of the conductor is zero. Consequently, the entire charge $Q$ must reside on the outer surface of the conductor.
$(b)$ The outer surface of conductor $A$ initially has a charge $Q$. When another conductor $B$ with charge $q$ is inserted into the cavity and insulated from $A$, an equal and opposite charge $-q$ is induced on the inner surface of the cavity of $A$ due to electrostatic induction. To conserve the total charge of the isolated conductor $A$, a charge $+q$ must be induced on its outer surface. Thus, the total charge on the outer surface of conductor $A$ becomes $Q+q$.
$(c)$ $A$ sensitive instrument can be shielded from strong electrostatic fields in its environment by enclosing it fully inside a metallic box or cage. This phenomenon is known as electrostatic shielding, where the electric field inside a closed metallic conductor is always zero.

(N/A) The charge placed at the centre of the shell is $+q$. Due to electrostatic induction,a charge of magnitude $-q$ is induced on the inner surface of the shell. Therefore,the total charge on the inner surface is $-q$.
The surface charge density on the inner surface is given by:
$\sigma_{1} = \frac{\text{Total charge}}{\text{Inner surface area}} = \frac{-q}{4 \pi r_{1}^{2}}$
$A$ charge of $+q$ is induced on the outer surface of the shell. Since the shell already has a charge $Q$,the total charge on the outer surface becomes $Q + q$. The surface charge density on the outer surface is:
$\sigma_{2} = \frac{\text{Total charge}}{\text{Outer surface area}} = \frac{Q + q}{4 \pi r_{2}^{2}}$
$(b)$ Yes,the electric field inside a cavity (with no charge) is zero,even if the shell is not spherical and has an irregular shape.
This is because if we consider a closed loop such that a part of it is inside the cavity along a field line and the rest is inside the conductor,the net work done by the electric field in moving a test charge along this closed loop must be zero. Since the electric field inside the conducting material is zero,the field inside the cavity must also be zero to satisfy the conservative nature of the electrostatic field.

(B) Let $a$ and $b$ be the radii of two conducting spheres $A$ and $B$,respectively. Let $Q_A$ and $Q_B$ be the charges on them,and $V_A$ and $V_B$ be their potentials.
Since the spheres are connected by a wire,they reach the same potential,so $V_A = V_B = V$.
The potential of a conducting sphere is given by $V = \frac{Q}{4 \pi \epsilon_0 R}$.
Thus,$V = \frac{Q_A}{4 \pi \epsilon_0 a} = \frac{Q_B}{4 \pi \epsilon_0 b}$,which implies $\frac{Q_A}{Q_B} = \frac{a}{b}$.
The electric field at the surface of a sphere is $E = \frac{Q}{4 \pi \epsilon_0 R^2}$.
The ratio of the electric fields is $\frac{E_A}{E_B} = \frac{Q_A / (4 \pi \epsilon_0 a^2)}{Q_B / (4 \pi \epsilon_0 b^2)} = \frac{Q_A}{Q_B} \times \frac{b^2}{a^2}$.
Substituting $\frac{Q_A}{Q_B} = \frac{a}{b}$,we get $\frac{E_A}{E_B} = \frac{a}{b} \times \frac{b^2}{a^2} = \frac{b}{a}$.
Since surface charge density $\sigma = \frac{Q}{4 \pi R^2} = \epsilon_0 E$,we have $\sigma \propto E$. For a conductor,the potential is constant throughout. $A$ sharp point has a very small radius of curvature $(R \to 0)$,which implies a very high electric field $(E \propto 1/R)$ and consequently a very high surface charge density $\sigma$ at that point.

(N/A) When we bring a charged body in contact with the earth,all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor. This process of sharing the charges with the earth is called grounding or earthing.
$A$ thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply.
The electric wiring in our houses has three wires: $live$,$neutral$,and $earth$.
The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate.
Metallic bodies of electric appliances such as electric iron,refrigerator,and $TV$ are connected to the earth wire. When any fault occurs or a live wire touches the metallic body,the charge flows to the earth without damaging the appliance and without causing any injury to humans.
Earthing provides a safety measure for the human body,electrical circuits,and appliances.

(A) For a conductor in electrostatic equilibrium,the electric field inside the material is zero. According to Gauss's Law,$\oint E \cdot dA = q_{enclosed} / \epsilon_0$. Since $E = 0$ inside the conductor,the net charge enclosed within any volume inside the conductor must be zero. Therefore,any excess or induced charge must reside entirely on the outer surface of the conductor.

(N/A) In metallic conductors, the charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move.
These electrons are free from their parent atoms but are not free to leave the metal. Hence, these free electrons form a kind of 'gas'.
Free electrons collide with each other and with the ions, moving randomly in different directions.
When placed in an external electric field $\vec{E}$, they drift against the direction of the field and get deposited on one surface, while an equal amount of positive charge is left on the opposite surface. This is shown in the figure.
The induced charges produce an internal electric field $\vec{E}_{in}$ inside the conductor in the direction opposite to the external electric field.
When the external electric field and the internal electric field become equal in magnitude, the net electric field inside the conductor becomes zero, and the charges stop depositing on the surface.

(N/A) The important results regarding the electrostatics of conductors are as follows:
$(1)$ The electrostatic field inside a conductor is zero.
$(2)$ On the outer surface of a charged conductor,the electric field at every point is perpendicular to the surface.
$(3)$ The interior of a conductor can have no excess charge in the static situation; any excess charge resides entirely on its surface.
$(4)$ The electrostatic potential is constant throughout the volume of the conductor and has the same value on its surface as it does inside.
$(5)$ The electric field at the surface of a charged conductor is given by $\vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n}$,where $\sigma$ is the surface charge density,$\epsilon_{0}$ is the permittivity of free space,and $\hat{n}$ is the unit vector normal to the surface in the outward direction.
$(6)$ The electric field inside the cavity of a conductor is zero,which leads to the phenomenon of electrostatic shielding.

(N/A) In the static situation,when there is no current inside or on the surface of the conductor,the electric field is zero everywhere inside the conductor.
$A$ conductor contains free electrons. If there were an electric field inside,these free charge carriers would experience a force $(F = qE)$ and would drift,creating a current.
In the static situation,the free charges redistribute themselves on the surface of the conductor such that the internal electric field produced by these charges exactly cancels the external applied electric field. Consequently,the net electric field becomes zero everywhere inside the conductor.

If the electric field $\vec{E}$ were not normal to the surface,it would have a non-zero tangential component along the surface.
This tangential component would exert a force on the free charges present on the conductor's surface,causing them to move.
Since the conductor is in an electrostatic (stable) state,there can be no net motion of charges.
Therefore,the tangential component of the electric field must be zero.
Thus,the electrostatic field at the surface of a charged conductor must be normal to the surface at every point,given by $\vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n}$,where $\sigma$ is the surface charge density and $\hat{n}$ is the unit normal vector.

(N/A) neutral conductor has equal amounts of positive and negative charges in every small volume or surface element.
When the conductor is charged,the excess charge can reside only on the surface in the static situation.
Let us consider a Gaussian surface inside the conductor and close to the surface.
At all points inside the conductor $\overrightarrow{E} = 0$,hence from $\phi_{E} = \oint \overrightarrow{E} \cdot d\overrightarrow{S}$,we get $\phi_{E} = 0$.
According to Gauss's law,$\phi_{E} = \frac{q}{\epsilon_{0}}$.
Since $\phi_{E} = 0$,it follows that $q = 0$.
Hence,there is no net charge at any point inside the conductor,and any excess charge must reside on the surface.

(N/A) In electrostatic equilibrium,the electric field $\vec{E}$ inside a conductor is zero.
Since the electric field is the negative gradient of the potential $(\vec{E} = -\nabla V)$,if $\vec{E} = 0$,then $\nabla V = 0$,which implies that the potential $V$ is constant throughout the volume of the conductor.
Furthermore,at the surface of the conductor,the electric field must be perpendicular to the surface. If there were a tangential component of the electric field,charges would move along the surface until the tangential component becomes zero.
Since there is no tangential component of the electric field on the surface,no work is done in moving a test charge between any two points on the surface. Therefore,the potential on the surface is the same as the potential inside the conductor.

(N/A) Electrostatic shielding is the phenomenon of protecting a certain region of space from an external electric field by surrounding it with a conductor.
Consider a conductor with a cavity as shown in the figure. Regardless of the size and shape of the cavity,if there are no charges inside it,the electric field inside the cavity remains zero.
When a conductor with a cavity is placed in an external electric field,the charges redistribute themselves on the outer surface such that the electric field inside the material of the conductor and within the cavity becomes zero.
Even if the conductor is charged,all excess charges reside only on the outer surface of the conductor. Thus,any cavity in a conductor remains shielded from external electric influences. This is known as electrostatic shielding.
$A$ practical example is sitting inside a car during a thunderstorm. If lightning strikes the car,the metal body of the car acts as an electrostatic shield,ensuring that the electric field inside remains zero,thus keeping the occupants safe.

(N/A) Consider a conductor with a cavity as shown in the figure.
There are no charges inside the cavity.
Regardless of the size and shape of the cavity, if a conductor with a cavity is placed in an external electric field, the electric field inside the cavity remains zero.
If the conductor is charged or charges are induced on a neutral conductor by an external field, all charges reside only on the outer surface of the conductor.
Any cavity in a conductor remains shielded from external electric influence. This phenomenon is known as electrostatic shielding.
For example, when we are in a car during a thunderstorm with lightning, we should close all the doors of the car. If lightning strikes the car (or a live current-carrying wire falls on the car), electrostatic shielding develops on the outer surface of the car, and we remain safe inside.

(N/A) Since the charge in the spherical cavity is $+Q$,the charge induced on the inner surface of the metallic spherical shell will be $-Q$ to ensure the electric field inside the conductor is zero.
Because the shell is metallic and neutral,the total charge on the shell must remain zero. Therefore,a charge of $+Q$ is induced on the outer surface to compensate for the $-Q$ charge on the inner surface.
The surface charge density $\sigma$ is defined as charge per unit area,$\sigma = \frac{q}{A}$.
$(i)$ For the inner surface with radius $R_1$,the surface charge density is $\sigma_1 = \frac{-Q}{4 \pi R_1^2}$.
$(ii)$ For the outer surface with radius $R_2$,the surface charge density is $\sigma_2 = \frac{+Q}{4 \pi R_2^2}$.

(N/A) In any neutral atom,the number of electrons and protons are equal,and they are bound together to form a stable,electrically neutral system.
Electrostatic fields in a macroscopic conductor arise due to the presence of net excess charges. According to the properties of conductors in electrostatic equilibrium,any excess charge resides only on the outer surface of the conductor.
Since there is no net excess charge inside the bulk of an isolated conductor,the internal electrostatic field is zero. The microscopic fields between protons and electrons within individual atoms cancel out on a macroscopic scale,resulting in no net macroscopic electric field inside the conductor.

(N/A) Principle: The Van de Graaff generator is based on two electrostatic principles:
$1$. The property that charge given to a hollow conductor is transferred to its outer surface and spreads uniformly over it.
$2$. The property that the electric potential of a conductor is higher when it is smaller in size.
Mathematical derivation:
Let a charge $Q$ be placed on a spherical shell of radius $R$. The potential at the surface is $V(R) = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{R}$.
If a small sphere of radius $r$ carrying charge $q$ is placed inside the shell,the total potential at the surface of the small sphere is $V(r) = \frac{1}{4 \pi \epsilon_{0}} \left( \frac{Q}{R} + \frac{q}{r} \right)$ and at the surface of the large shell is $V(R) = \frac{1}{4 \pi \epsilon_{0}} \left( \frac{Q}{R} + \frac{q}{R} \right)$.
The potential difference is $V(r) - V(R) = \frac{q}{4 \pi \epsilon_{0}} \left( \frac{1}{r} - \frac{1}{R} \right)$.
Since $r < R$,the term $(\frac{1}{r} - \frac{1}{R})$ is positive,meaning the inner sphere is at a higher potential than the outer shell. When connected,charge flows from the inner sphere to the outer shell,allowing the accumulation of a very high potential on the outer shell.

(N/A) Construction: $A$ Van de Graaff generator consists of a large spherical conducting shell supported on an insulating column. $A$ long,narrow,endless belt made of an insulating material like rubber or silk is wound around two pulleys.
As shown in the diagram,one pulley is at the ground level and the other is at the center of the shell. The belt is kept in continuous motion by a motor driving the lower pulley.
$A$ metal brush at the ground level sprays positive charge onto the belt,which is then carried upwards by the belt.
The positive charge reaching the top is transferred to a second metal brush connected to the inner side of the shell.
Consequently,the positive charge spreads uniformly over the outer surface of the shell,creating a high potential.
In this way,a potential difference of $6$ to $8$ million volts relative to the ground can be generated.
Uses: It is used to accelerate charged particles like protons,deuterons,and alpha particles to very high energies,which are then used to probe the structure of atomic nuclei.

(B) Both spheres are at the same potential,so $V_1 = V_2$.
Using the formula for potential $V = \frac{kq}{R}$,we have $\frac{kq_1}{R_1} = \frac{kq_2}{R_2}$.
Since the charge $q = \sigma A = \sigma (4\pi R^2)$,we substitute this into the potential equation:
$\frac{k(\sigma_1 4\pi R_1^2)}{R_1} = \frac{k(\sigma_2 4\pi R_2^2)}{R_2}$.
Simplifying this,we get $\sigma_1 R_1 = \sigma_2 R_2$.
Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.
Since $R_1 > R_2$,it follows that $\frac{R_2}{R_1} < 1$,which implies $\sigma_1 < \sigma_2$.
Thus,the charge density of the smaller sphere $(R_2)$ is greater than the charge density of the larger sphere $(R_1)$.

(A) Yes,there can be a potential difference if the sizes or shapes of the conductors are different.
The capacitance $C$ of a conductor is defined by the relation $C = \frac{Q}{V}$,where $Q$ is the charge on the conductor and $V$ is its electric potential.
Rearranging this formula,we get $V = \frac{Q}{C}$.
For a given charge $Q$,the potential $V$ is inversely proportional to the capacitance $C$ $(V \propto \frac{1}{C})$.
Since capacitance depends on the geometry (size and shape) of the conductor,two adjacent conductors with the same charge but different dimensions will have different capacitances,and consequently,different electric potentials.

(N/A) The electric field $E = -\frac{dV}{dr}$ indicates that the electric potential $V$ decreases in the direction of the electric field.
Consider a path starting from the charged conductor and moving towards the uncharged conductor along the direction of the electric field lines. Since the field originates from the charged body,the potential $V$ must decrease along this path. Thus,the potential of the uncharged body $V_u$ is less than the potential of the charged body $V_c$ (assuming $V_c > 0$).
Next,consider a path from the uncharged conductor extending towards infinity. Since the electric field lines continue to point away from the charged system towards infinity,the potential must continue to decrease along this path. By definition,the potential at infinity is $V_{\infty} = 0$.
Therefore,the potential of the uncharged body $V_u$ satisfies the condition $V_{\infty} < V_u < V_c$. This confirms that the uncharged body is intermediate in potential between the charged body and infinity.

(B) When two conductors are connected by a wire,charge flows until their potentials become equal.
Let $V_{1}$ and $V_{2}$ be the potentials of the spheres. Since they are connected,$V_{1} = V_{2}$.
The potential of a charged spherical conductor is given by $V = \frac{kQ}{R}$.
Thus,$\frac{kQ_{1}}{R_{1}} = \frac{kQ_{2}}{R_{2}}$,which implies $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{A} = \frac{Q}{4\pi R^{2}}$.
Therefore,the ratio of surface charge densities is:
$\frac{\sigma_{1}}{\sigma_{2}} = \frac{Q_{1} / (4\pi R_{1}^{2})}{Q_{2} / (4\pi R_{2}^{2})} = \frac{Q_{1}}{Q_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}$.
Substituting $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$,we get:
$\frac{\sigma_{1}}{\sigma_{2}} = \frac{R_{1}}{R_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}$.

(B) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_A = V_B$.
Since the potential of a spherical conductor is $V = \frac{KQ}{R}$,we have $\frac{KQ_A}{R_A} = \frac{KQ_B}{R_B}$.
This implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{5 \ mm}{10 \ mm} = \frac{1}{2}$.
The electric field at the surface of a sphere is given by $E = \frac{KQ}{R^2}$.
Therefore,the ratio of the electric fields is $\frac{E_A}{E_B} = \frac{KQ_A / R_A^2}{KQ_B / R_B^2} = \frac{Q_A}{Q_B} \times \left(\frac{R_B}{R_A}\right)^2$.
Substituting the ratio of charges: $\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}\right) \times \left(\frac{R_B}{R_A}\right)^2 = \frac{R_B}{R_A}$.
Given $R_A = 5 \ mm$ and $R_B = 10 \ mm$,we get $\frac{E_A}{E_B} = \frac{10}{5} = \frac{2}{1}$.

(A) Statement $I$ is true: In electrostatic equilibrium,the interior of a conductor has no net electric field $(E = 0)$. Since $E = -dV/dr$,if $E = 0$,then the potential $V$ must be constant throughout the volume and on the surface of the conductor.
Statement $II$ is true: Since the conductor is an equipotential surface,any electric field component tangential to the surface would cause charges to move along the surface. To maintain equilibrium,the electric field must have no tangential component,meaning it must be strictly perpendicular to the surface at every point.

(D) When a point charge $+Q$ is placed inside the hollow conducting spherical shell,it induces a charge of $-Q$ on the inner surface of the shell due to electrostatic induction.
To maintain the neutrality of the shell,a charge of $+Q$ appears on the outer surface of the shell.
When the outer surface of the shell is connected to the earth (grounded),the positive charge $+Q$ on the outer surface flows to the earth,leaving the outer surface neutral.
According to the property of electrostatic shielding,the electric field produced by the charge $+Q$ inside the shell and the induced charge $-Q$ on the inner surface is confined within the cavity of the shell.
Therefore,the net electric field at any point outside the conducting shell is zero.
Since the electric field at point $P$ (which is outside the shell) is zero,the electrostatic force on a test charge $+q$ placed at $P$ is also zero.

(A) Since spheres $A$ and $B$ are connected,they act as a single conductor at the same potential. Let the charge on the combined system $(A+B)$ be $q$ and the charge on sphere $C$ be $Q_C$.
The potential of the system $(A+B)$ is given by:
$V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{b} + \frac{Q_C}{c} \right) = V$
Since sphere $C$ is grounded,its potential is zero:
$V_C = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{c} + \frac{Q_C}{c} \right) = 0$
From the second equation,we get $q + Q_C = 0$,which implies $q = -Q_C$.
Substituting $q = -Q_C$ into the first equation:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{-Q_C}{b} + \frac{Q_C}{c} \right)$
$V = \frac{Q_C}{4 \pi \varepsilon_0} \left( \frac{1}{c} - \frac{1}{b} \right) = \frac{Q_C}{4 \pi \varepsilon_0} \left( \frac{b-c}{bc} \right)$
Solving for $Q_C$:
$Q_C = 4 \pi \varepsilon_0 V \left( \frac{bc}{b-c} \right) = -4 \pi \varepsilon_0 V \left( \frac{bc}{c-b} \right)$

(D) According to the properties of a conducting shell,the electric field inside the cavity is determined solely by the charge $+q$ placed within the cavity.
The conducting material of the shell redistributes its own charges to shield the cavity from any external electric fields (such as the field produced by $+Q$ outside the shell).
Specifically,the charge $+q$ inside induces a charge $-q$ on the inner surface of the shell,and the external field $+Q$ induces charges on the outer surface of the shell,but these external effects do not penetrate the cavity.
Therefore,the electric field in the hollow cavity is only that due to $+q$.

(D) The potential at any point on the surface of a grounded conducting sphere is always zero $(V = 0)$.
The total induced charge $q_{\text{induced}}$ on a grounded sphere due to an external charge distribution can be found using the method of images or by considering the potential at the center of the sphere.
For a dipole placed at a distance $D$ from the center of the sphere,the potential at the center of the sphere due to the dipole is given by $V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}$.
Since the dipole moment vector $\vec{p}$ is perpendicular to the line joining the center of the dipole and the center of the sphere (the position vector $\vec{r}$),the dot product $\vec{p} \cdot \vec{r} = 0$.
Therefore,the potential at the center of the sphere due to the dipole is zero. Since the sphere is grounded,the total induced charge on the sphere must be zero to maintain the potential at zero.

(B) When a conducting soap bubble is given a charge (whether positive or negative),the charge distributes itself uniformly over the outer surface of the bubble due to electrostatic repulsion between like charges.
This distribution of charge creates an outward electrostatic pressure on the surface of the bubble.
This outward pressure acts in addition to the existing internal gas pressure,causing the bubble to expand until a new equilibrium is reached between the electrostatic pressure,the internal gas pressure,and the surface tension of the soap film.

(C) Let the charge on the inner surface of the outermost shell (radius $r_3$) be $q'$.
Consider a Gaussian surface inside the material of the outermost shell. Since the electric field inside a conductor in electrostatic equilibrium is zero,the net charge enclosed by this Gaussian surface must be zero.
The total charge enclosed by this surface is the sum of the charge on the innermost shell $(q_1)$,the charge on the middle shell $(0)$,and the charge on the inner surface of the outermost shell $(q')$.
Therefore,$q_1 + 0 + q' = 0$.
This gives $q' = -q_1$.

(D) When a neutral conducting sphere is placed in a uniform electric field between the plates of a parallel plate capacitor,the charges on the sphere redistribute due to electrostatic induction.
Negative charges accumulate on the side facing the positively charged plate $A$,and positive charges accumulate on the side facing the negatively charged plate $B$.
The electric field $E$ between the plates is uniform and directed from plate $A$ to plate $B$.
The force on the induced negative charge is $F_- = -qE$ (towards plate $A$),and the force on the induced positive charge is $F_+ = +qE$ (towards plate $B$).
Since the sphere is neutral and the electric field is uniform,the magnitude of the induced charges is equal,and the forces $F_-$ and $F_+$ are equal in magnitude and opposite in direction.
Therefore,the net electrostatic force on the sphere is $F_{net} = F_+ + F_- = 0$.
Thus,the net force on the sphere is zero.

(B) Let the charge on the inner sphere be $q$. Since the inner sphere is earthed,its potential $V$ must be zero.
The potential at the surface of the inner sphere is due to its own charge $q$ and the charge $Q = 6 \, \mu C$ on the outer shell.
The potential $V$ at the inner sphere (radius $r_1 = 1 \, m$) is given by:
$V = \frac{k q}{r_1} + \frac{k Q}{r_2} = 0$
Substituting the given values $r_1 = 1 \, m$,$r_2 = 3 \, m$,and $Q = 6 \, \mu C$:
$\frac{k q}{1} + \frac{k (6)}{3} = 0$
$k q + 2k = 0$
$q = -2 \, \mu C$
Thus,the charge on the inner sphere is $-2 \, \mu C$.

(D) When a conducting sphere is earthed, its potential becomes $0$.
Let the charge on the inner surface of the hollow sphere be $q_{in}$ and the charge on the outer surface be $q_{out}$.
By the property of induction, $q_{in} = -q$.
Since the outer sphere is earthed, its potential $V$ is the sum of potentials due to the inner sphere (charge $q$) and the hollow sphere (charges $q_{in}$ and $q_{out}$).
The potential at the surface of the outer sphere (radius $2R$) is given by:
$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{2R} + \frac{q_{in}}{2R} + \frac{q_{out}}{2R} \right) = 0$.
Substituting $q_{in} = -q$:
$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{2R} - \frac{q}{2R} + \frac{q_{out}}{2R} \right) = 0$.
This simplifies to $\frac{q_{out}}{2R} = 0$, which implies $q_{out} = 0$.
Therefore, the charge on the outer surface is $0$.

(B) Inside a metallic conductor,the net electric field must be zero in electrostatic equilibrium.
Let $\vec{E}_{ext}$ be the electric field at the centre of the shell due to the external point charge $q$.
Let $\vec{E}_{ind}$ be the electric field at the centre of the shell due to the induced charges on the surface of the shell.
Since the net electric field at the centre is zero,we have $\vec{E}_{ext} + \vec{E}_{ind} = 0$,which implies $\vec{E}_{ind} = -\vec{E}_{ext}$.
The magnitude of the electric field due to the point charge $q$ at the centre of the shell (which is at a distance $R + 2R = 3R$ from the charge) is given by $E_{ext} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{(3R)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{9R^2}$.
Therefore,the magnitude of the electric field due to the induced charge is $|\vec{E}_{ind}| = |\vec{E}_{ext}| = \frac{1}{4 \pi \varepsilon_0} \frac{q}{9R^2}$.

(A) The electric potential $V$ of a charged conducting sphere of radius $r$ carrying charge $q$ is given by $V = \frac{kq}{r}$,where $k = \frac{1}{4\pi\epsilon_0}$.
For sphere $A$ with radius $a$ and charge $q = 100 \mu C$,the potential is $V_A = \frac{k(100)}{a}$.
For sphere $B$ with radius $b$ and charge $q = 100 \mu C$,the potential is $V_B = \frac{k(100)}{b}$.
Given that $a < b$,it follows that $\frac{1}{a} > \frac{1}{b}$.
Therefore,$V_A > V_B$.
Since charge flows from a region of higher potential to a region of lower potential,charge will flow from sphere $A$ to sphere $B$ until their potentials become equal.

(D) Initial charges on the spheres are:
$Q_1 = \sigma(4\pi R^2) = 4\pi R^2\sigma$
$Q_2 = \sigma(4\pi(2R)^2) = 16\pi R^2\sigma$
Total charge $Q = Q_1 + Q_2 = 20\pi R^2\sigma$.
When connected by a wire,the potentials become equal:
$V_1 = V_2 \implies \frac{kQ_1^{\prime}}{R} = \frac{kQ_2^{\prime}}{2R} \implies Q_2^{\prime} = 2Q_1^{\prime}$.
Since charge is conserved,$Q_1^{\prime} + Q_2^{\prime} = 20\pi R^2\sigma$.
Substituting $Q_1^{\prime} = \frac{Q_2^{\prime}}{2}$,we get $\frac{Q_2^{\prime}}{2} + Q_2^{\prime} = 20\pi R^2\sigma \implies \frac{3}{2}Q_2^{\prime} = 20\pi R^2\sigma \implies Q_2^{\prime} = \frac{40}{3}\pi R^2\sigma$.
The new charge density of the bigger sphere is $\sigma^{\prime} = \frac{Q_2^{\prime}}{4\pi(2R)^2} = \frac{Q_2^{\prime}}{16\pi R^2}$.
Substituting $Q_2^{\prime}$,we get $\sigma^{\prime} = \frac{40\pi R^2\sigma}{3 \cdot 16\pi R^2} = \frac{40\sigma}{48} = \frac{5}{6}\sigma$.
Therefore,$\frac{\sigma^{\prime}}{\sigma} = \frac{5}{6}$.

(B) Region $I$ $(r < a)$: The electric field is due to the point charge $Q$ at the center. By Gauss's Law,$E_I = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \neq 0$.
Region $II$ $(a < r < b)$: This region lies within the material of the conducting shell. In electrostatic equilibrium,the electric field inside the material of a conductor is always zero $(E_{II} = 0)$.
Region $III$ $(r > b)$: The charge $Q$ induces a charge $-Q$ on the inner surface $(r=a)$ and $+Q$ on the outer surface $(r=b)$. The total charge enclosed by a Gaussian surface of radius $r > b$ is $Q + (-Q) + Q = Q$. Thus,$E_{III} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \neq 0$.

(C) When two conducting spheres are connected by a conducting wire,charge flows until they reach the same electric potential.
Let $q_1$ and $q_2$ be the charges on the spheres of radii $a$ and $b$ respectively.
The potential $V$ at the surface of a conducting sphere is given by $V = \frac{Kq}{r}$,where $K = \frac{1}{4\pi\epsilon_0}$.
Since the potentials are equal,we have:
$V_1 = V_2$
$\frac{Kq_1}{a} = \frac{Kq_2}{b}$
Rearranging the terms to find the ratio of charges:
$\frac{q_1}{q_2} = \frac{a}{b}$
Thus,the ratio of the charges is $\frac{a}{b}$.

(B) When two conductors are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_A$ and $Q_B$.
Since $V_A = V_B$,we have $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$,which implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$.
Since $R_A > R_B$,it follows that $Q_A > Q_B$. Thus,$(B)$ is correct.
For a spherical shell,the electric field inside is always zero due to Gauss's Law. Thus,$(A)$ is correct.
The surface charge density is $\sigma = \frac{Q}{4\pi R^2}$. Therefore,$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi R_A^2} \cdot \frac{4\pi R_B^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_A}{R_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_B}{R_A}$. Thus,$(C)$ is correct.
The electric field on the surface is $E = \frac{\sigma}{\epsilon_0}$. Since $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A} < 1$,we have $\sigma_A < \sigma_B$,which implies $E_A < E_B$. Thus,$(D)$ is correct.
Therefore,all options $(A, B, C, D)$ are correct.

(A) The phenomenon described is known as electrostatic shielding. According to the properties of conductors in electrostatic equilibrium,the electric field inside a closed metallic cavity is always zero,regardless of the external electric field or charge distribution. When lightning strikes an aircraft,the metal body acts as a Faraday cage,ensuring that the electric field inside remains zero and the occupants remain safe. Therefore,Assertion $(A)$ is correct because it describes the protective effect,and Reason $(R)$ is the correct scientific explanation for this phenomenon.

(D) The potential $V$ of a conducting sphere of radius $r$ with surface charge density $\sigma$ is given by $V = \frac{\sigma r}{\varepsilon_0}$.
When two conducting spheres are brought into contact,charge flows between them until their potentials become equal,i.e.,$V_1 = V_2$.
Substituting the expression for potential,we get $\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$.
This simplifies to $\sigma_1 r_1 = \sigma_2 r_2$.
Given $r_1 = R$ and $r_2 = 3R$,we have $\sigma_1 R = \sigma_2 (3R)$.
Therefore,the ratio $\frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$.