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

(C) Before closing the switch,the potential at point $A$ (on the surface of the inner shell of radius $a$) is due to both shells:
$V_A = \frac{kQ}{a} + \frac{k(2Q)}{2a} = \frac{kQ}{a} + \frac{kQ}{a} = \frac{2kQ}{a} = V$.
Thus,$\frac{kQ}{a} = \frac{V}{2}$.
After closing the switch,the two shells are connected,so the total charge $Q + 2Q = 3Q$ redistributes such that both shells are at the same potential. The potential at the surface of the inner shell (and outer shell) becomes:
$V'_A = \frac{k(Q + 2Q)}{2a} = \frac{3kQ}{2a}$.
Substituting $\frac{kQ}{a} = \frac{V}{2}$ into the expression:
$V'_A = \frac{3}{2} \times \left(\frac{kQ}{a}\right) = \frac{3}{2} \times \frac{V}{2} = \frac{3V}{4}$.

(D) The electric field at the surface of a conductor is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density. For a conductor of a fixed volume,the surface charge density $\sigma$ is higher at regions with smaller radii of curvature (sharp points).
When the electric field at the surface exceeds the dielectric strength of the surrounding air (approximately $3 \times 10^6 \ V/m$),the air undergoes dielectric breakdown,causing charge leakage (corona discharge).
Since sharp points reach this critical electric field value at lower total charges compared to smooth surfaces,the maximum charge a conductor can hold depends on its shape. Thus,the Reason correctly explains why the shape affects the maximum charge.

(C) When two conductors are connected by a thin conducting wire,charge flows until they reach the same potential $V$.
Since $V_1 = V_2$,we have $\frac{k q_1}{R} = \frac{k q_2}{r}$,which implies $\frac{q_1}{q_2} = \frac{R}{r}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi R^2}$.
Therefore,the ratio of the surface charge densities is:
$\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi R^2)}{q_2 / (4 \pi r^2)} = \left( \frac{q_1}{q_2} \right) \left( \frac{r^2}{R^2} \right)$.
Substituting $\frac{q_1}{q_2} = \frac{R}{r}$,we get:
$\frac{\sigma_1}{\sigma_2} = \left( \frac{R}{r} \right) \left( \frac{r^2}{R^2} \right) = \frac{r}{R}$.
Thus,the ratio is $r: R$.

(C) For a conductor in electrostatic equilibrium,the electric field just outside the surface must be perpendicular to the surface at every point. If it had a tangential component,charges would experience a force and move along the surface,which contradicts the static condition. Therefore,the statement that the field must be tangential is '$WRONG$'.

(B) Since the spheres are joined by a metal wire,their potentials will be the same.
Let the radius of the smaller sphere be $r_1$ and the radius of the larger sphere be $r_2 = 4r_1$.
Since the potentials are equal,$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2}$.
Therefore,$\frac{q_2}{q_1} = \frac{r_2}{r_1} = 4$.
The electric field near the surface of the spheres is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}$.
Thus,$\frac{E_2}{E_1} = \frac{q_2}{q_1} \cdot \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{r_2}{r_1}\right) \cdot \left(\frac{r_1}{r_2}\right)^2 = \frac{r_1}{r_2}$.
Substituting $r_2 = 4r_1$,we get $\frac{E_2}{E_1} = \frac{r_1}{4r_1} = \frac{1}{4}$.
So,the electric field near the larger sphere is a quarter of that near the smaller sphere.

(B) When two isolated metallic spheres are brought into contact,charge flows between them until their potentials become equal.
Let $r_A$ and $r_B$ be the radii of spheres $A$ and $B$ respectively,and $q_A$ and $q_B$ be their final charges.
Given that $r_B = \frac{r_A}{2}$,which implies $r_A = 2r_B$.
Since the potentials are equal,$V_A = V_B$.
Using the formula for potential $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$,we have:
$\frac{1}{4 \pi \varepsilon_0} \frac{q_A}{r_A} = \frac{1}{4 \pi \varepsilon_0} \frac{q_B}{r_B}$
$\frac{q_A}{q_B} = \frac{r_A}{r_B}$
Substituting $r_A = 2r_B$:
$\frac{q_A}{q_B} = \frac{2r_B}{r_B} = \frac{2}{1}$
Thus,the ratio of the charges on spheres $A$ and $B$ is $2: 1$.

(A) When two charged conducting spheres are connected by a wire,charge flows until their electric potentials become equal.
Therefore,$V_1 = V_2$.
The potential of a charged sphere is given by $V = \frac{KQ}{R}$.
Thus,$\frac{KQ_1}{R_1} = \frac{KQ_2}{R_2}$.
The electric field at the surface of a sphere is $E = \frac{KQ}{R^2}$.
We can rewrite the potential equation as $\frac{KQ_1}{R_1^2} \cdot R_1 = \frac{KQ_2}{R_2^2} \cdot R_2$.
Substituting $E = \frac{KQ}{R^2}$,we get $E_1 R_1 = E_2 R_2$.
Therefore,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{R_2}{R_1}$.

(C) Let the potentials of the spheres with radii $a$ and $b$ be $V_a$ and $V_b$ respectively,and the charges on them be $Q_a$ and $Q_b$.
Since the two spheres are connected by a conducting wire,they must be at the same potential:
$V_a = V_b$
$\frac{k Q_a}{a} = \frac{k Q_b}{b}$
$\frac{Q_a}{Q_b} = \frac{a}{b}$
Using the property of ratios,$\frac{Q_a}{Q_a + Q_b} = \frac{a}{a + b}$.
Since the total charge $Q_a + Q_b = Q$,we have:
$Q_a = \frac{a Q}{a + b}$ and $Q_b = \frac{b Q}{a + b}$.
Now,calculating the potential $V$ of each sphere:
$V = V_a = V_b = \frac{k Q_a}{a} = \frac{k}{a} \left( \frac{a Q}{a + b} \right) = \frac{k Q}{a + b}$.

(B) In a static situation,the interior of a conductor cannot have any excess charge. According to Gauss's Law,if there were an excess charge $q$ inside a closed surface,there would be a net electric flux $\phi = q/\epsilon_0$ through the surface. However,since the electric field inside a conductor is zero,the flux must be zero,implying that the net charge inside any volume of the conductor is zero. Therefore,any excess charge must reside on the surface of the conductor. Thus,option $B$ is incorrect.

(C) In electrostatic equilibrium,the electric field inside a conductor is zero. According to Gauss's law,any excess charge placed on a conductor must reside entirely on its surface to minimize mutual repulsion.
The electric field just outside the surface of a charged conductor is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma$ is the surface charge density.
The electric potential inside a charged conductor is constant and equal to the potential at its surface,not zero.
The net electric field is always perpendicular (normal) to the surface of the conductor,not tangential.
Therefore,the correct statement is that any excess charge resides on the surface of the conductor.

(B) Let the plates be numbered $1, 2, 3, 4, 5$ from left to right. Plate $2$ is $M$ and plate $4$ is $N$.
Since plates $1$ and $5$ are grounded,their potentials are zero. The electric field in the regions outside the system is zero.
Let the charges on the inner and outer surfaces of the plates be $q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8, q_9, q_{10}$.
Due to grounding,the outer surfaces of plates $1$ and $5$ have zero charge. The total charge on plate $M$ is $Q_1$ and on plate $N$ is $Q_2$.
Using the property that the electric field inside a conductor is zero,the potential difference between plates $M$ and $N$ is given by $V_M - V_N = E \cdot d$,where $E$ is the electric field between them.
For this configuration,the electric field between plates $M$ and $N$ is $E = \frac{Q_1 - Q_2}{2 \epsilon_0 A}$.
Thus,the potential difference is $\Delta V = E \cdot d = \frac{d(Q_1 - Q_2)}{2 \epsilon_0 A}$.
Therefore,the correct option is $B$.

(B) Let the charge on the inner shell after the switch is closed be $q$. Since the inner shell is connected to the ground,its potential becomes zero.
The potential at the inner shell of radius $R$ due to its own charge $q$ and the charge $Q$ on the outer shell of radius $2R$ is:
$V_{\text{inner}} = \frac{Kq}{R} + \frac{KQ}{2R} = 0$
Solving for $q$:
$\frac{Kq}{R} = -\frac{KQ}{2R}$
$q = -\frac{Q}{2}$
Thus,the potential of the inner shell becomes zero,making statement $(a)$ correct.
The charge on the inner shell is $-\frac{Q}{2}$,making statement $(b)$ wrong.
Therefore,$(a)$ is correct and $(b)$ is wrong.

(B) For a charged hollow sphere,the electric field $(E)$ inside the sphere is zero because there is no enclosed charge $(q_{enclosed} = 0)$.
According to the relation between electric field and potential,$E = -\frac{dV}{dr}$.
Since $E = 0$,it implies that $\frac{dV}{dr} = 0$,which means the potential $(V)$ must be constant throughout the interior of the sphere.
The value of this constant potential is equal to the potential at the surface of the sphere,given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.

(D) Since the inner sphere is grounded,its potential must be zero due to the charges on both the outer sphere and the inner sphere.
Let $r_1$ be the radius of the inner sphere and $r_2$ be the radius of the outer sphere.
The potential $V$ at the surface of the inner sphere is the sum of the potentials due to its own charge $q'$ and the charge $q$ on the outer sphere.
$V = \frac{k q'}{r_1} + \frac{k q}{r_2} = 0$
Where $k = \frac{1}{4 \pi \epsilon_0}$.
Rearranging the equation:
$\frac{k q'}{r_1} = -\frac{k q}{r_2}$
$q' = -\frac{r_1}{r_2} q$

(C) Let $q_1$ and $q_2$ be the final charges on sphere '$1$' and sphere '$2$' respectively.
By the principle of conservation of charge,$q_1 + q_2 = q$.
When connected by a conducting wire,the potentials of the two spheres become equal,so $V_1 = V_2$.
Using the formula for potential $V = \frac{Kq}{R}$,we have $\frac{Kq_1}{R} = \frac{Kq_2}{3R}$.
This simplifies to $q_2 = 3q_1$.
Substituting this into the charge conservation equation: $q_1 + 3q_1 = q \Rightarrow 4q_1 = q \Rightarrow q_1 = \frac{q}{4}$ and $q_2 = \frac{3q}{4}$.
The surface charge density is given by $\sigma = \frac{q}{A} = \frac{q}{4\pi r^2}$.
Therefore,$\sigma_1 = \frac{q_1}{4\pi R^2} = \frac{q/4}{4\pi R^2} = \frac{q}{16\pi R^2}$.
And $\sigma_2 = \frac{q_2}{4\pi (3R)^2} = \frac{3q/4}{4\pi (9R^2)} = \frac{3q}{144\pi R^2} = \frac{q}{48\pi R^2}$.
The ratio is $\frac{\sigma_1}{\sigma_2} = \frac{q / 16\pi R^2}{q / 48\pi R^2} = \frac{48}{16} = 3$.

(C) Let the radii of the two spheres be $R_1 = 4R$ and $R_2 = 7R$. When they are in contact,they reach the same potential $V$.
Since $V = \frac{Kq}{R}$,we have $\frac{Kq_1}{R_1} = \frac{Kq_2}{R_2}$.
This implies $\frac{q_1}{4R} = \frac{q_2}{7R}$,so $\frac{q_1}{q_2} = \frac{4}{7}$.
The total charge $q = q_1 + q_2 = 8.8 \times 10^{-7} \text{ C}$.
Substituting $q_2 = \frac{7}{4}q_1$,we get $q_1 + \frac{7}{4}q_1 = 8.8 \times 10^{-7} \text{ C}$.
$\frac{11}{4}q_1 = 8.8 \times 10^{-7} \text{ C} \Rightarrow q_1 = \frac{4}{11} \times 8.8 \times 10^{-7} = 3.2 \times 10^{-7} \text{ C}$.
The potential at a distance $r = 60 \text{ m}$ from the smaller sphere is $V = \frac{Kq_1}{r}$.
$V = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{60} = \frac{28.8 \times 10^2}{60} = 48 \text{ V}$.

(A) Let the charges on the two plates be $Q_1 = 12 \mu C$ and $Q_2 = 6 \mu C$.
When two large conducting plates are placed parallel to each other,the charge on the outer surfaces is equal to $\frac{Q_1 + Q_2}{2}$ and the charge on the inner surfaces is equal to $\frac{Q_1 - Q_2}{2}$ and $\frac{Q_2 - Q_1}{2}$ respectively.
For surface $A$ (outer surface of first plate): $q_A = \frac{Q_1 + Q_2}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9 \mu C$.
For surface $B$ (inner surface of first plate): $q_B = \frac{Q_1 - Q_2}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3 \mu C$.
For surface $C$ (inner surface of second plate): $q_C = \frac{Q_2 - Q_1}{2} = \frac{6 - 12}{2} = \frac{-6}{2} = -3 \mu C$.
For surface $D$ (outer surface of second plate): $q_D = \frac{Q_1 + Q_2}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9 \mu C$.
Thus,the charge distribution on surfaces $A, B, C$ and $D$ is $9 \mu C, 3 \mu C, -3 \mu C$ and $9 \mu C$ respectively.

(C) When a charge $q_a$ is placed inside a cavity of a conducting sphere,it induces a charge $-q_a$ on the inner surface of the cavity and a charge $+q_a$ on the outer surface of the conducting sphere.
Similarly,placing $q_b$ in the other cavity induces $-q_b$ on its inner surface and $+q_b$ on the outer surface of the conducting sphere.
The total charge on the outer surface of the conducting sphere becomes $q_a + q_b$. Since the sphere is conducting,the electric field inside the bulk of the conductor is zero.
The charge $q_a$ experiences a force due to the field produced by the induced charge $-q_b$ on the inner surface of the second cavity and the field produced by the charge distribution on the outer surface of the sphere.
However,due to the electrostatic shielding property of the conductor,the electric field produced by the charges outside the cavity (including the outer surface charge and the other cavity's induced charge) is zero inside the cavity containing $q_a$.
Therefore,the charge $q_a$ only experiences the field produced by the charge $-q_b$ induced on the inner surface of its own cavity,which is zero at its center due to symmetry.
Thus,the net force on $q_a$ due to $q_b$ and the induced charges is zero.

(B) Inside a hollow charged metal sphere,the electric field intensity is always zero because there is no charge enclosed within the Gaussian surface inside the sphere.
$\therefore E = 0$.
However,the electric potential inside the sphere is constant and equal to the potential at its surface.
The potential on the surface of the hollow sphere is given by:
$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} = \frac{Q}{4 \pi \varepsilon_{0} R}$.
Thus,inside the sphere,the potential is $\frac{Q}{4 \pi \varepsilon_{0} R}$ and the electric field intensity is zero.

(C) Let the charge on the inner shell be $q^{\prime}$.
Since the inner shell is grounded,its electric potential must be zero.
The potential at the surface of the inner shell is due to its own charge $q^{\prime}$ and the charge $q$ on the outer shell.
The potential $V_{1}$ at the inner shell is given by $V_{1} = \frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}}$.
Setting $V_{1} = 0$,we get $\frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}} = 0$.
Solving for $q^{\prime}$,we get $\frac{q^{\prime}}{r_{1}} = -\frac{q}{r_{2}}$.
Therefore,$q^{\prime} = -\left(\frac{r_{1}}{r_{2}}\right) q$.

(D) When two conductors are connected by a wire,their potentials become equal $(V_1 = V_2)$.
For a conducting sphere of radius $R$ and charge $q$,the potential is $V = \frac{kq}{R}$.
Since the potentials are equal,we have $\frac{kq_1}{R_1} = \frac{kq_2}{R_2}$,which implies $\frac{q_1}{q_2} = \frac{R_1}{R_2}$.
The electric field on the surface of a sphere is $E = \frac{kq}{R^2}$.
Therefore,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{kq_1/R_1^2}{kq_2/R_2^2} = \frac{q_1}{q_2} \cdot \frac{R_2^2}{R_1^2}$.
Substituting $\frac{q_1}{q_2} = \frac{R_1}{R_2}$ into the equation,we get $\frac{E_1}{E_2} = \frac{R_1}{R_2} \cdot \frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}$.
Given $R_1 = 8 \text{ cm}$ and $R_2 = 18 \text{ cm}$,the ratio is $\frac{E_1}{E_2} = \frac{18}{8} = \frac{9}{4}$.

(A) is correct: In electrostatic equilibrium,$E = 0$ inside a conductor.
$B$ is incorrect: $E = \frac{\sigma}{\epsilon_0}$ at the surface,so it depends on $\sigma$.
$C$ is correct: Excess charge resides only on the surface of a conductor in static conditions.
$D$ is correct: Field lines must be perpendicular to the surface of a conductor to ensure no tangential force,which would otherwise cause current.
$E$ is incorrect: Potential is constant inside,not zero.
Thus,$A, C$,and $D$ are correct. Option $A$ is correct.