Electrostatics

The charge on the sphere after removing electrons is given by $q = ne$, where $n$ is the number of electrons removed and $e$ is the elementary charge ($1.6 \times 10^{-19}\ \text{C}$). Since electrons are removed, the sphere acquires a positive charge. Substituting the given values, $q = 10^{14} \times 1.6 \times 10^{-19}\ \text{C} = 1.6 \times 10^{-5}\ \text{C} = 16\ \mu\text{C}$.

The number of missing electrons responsible for the positive charge is given by $n = \frac{q}{e} = \frac{14.4 \times 10^{-19}\ \text{C}}{1.6 \times 10^{-19}\ \text{C}} = 9$. Since the charge is positive, the conductor has a deficiency or is short of 9 electrons.

An alpha particle consists of 2 protons and 2 neutrons.
The charge of a single proton is $e = 1.6 \times 10^{-19}\text{ C}$.
The net charge $q$ on the alpha particle is given by $q = 2e$.
Substituting the value of $e$, we get $q = 2 \times 1.6 \times 10^{-19}\text{ C}$.
Thus, the total charge is $3.2 \times 10^{-19}\text{ C}$.

All material bodies contain a large number of electrons and equal numbers of protons in their normal neutral state.
When a glass rod is rubbed against a silk cloth, friction causes some electrons to be transferred from the glass rod to the silk cloth.
The glass rod loses electrons and therefore becomes positively charged.
The silk cloth receives these electrons and becomes negatively charged.
Thus, the glass rod gives electrons to the silk.
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The earth acts as a large reservoir of electric charges and is a good conductor.
If electrons flow from the earth into a body during grounding, it indicates that there is a deficiency of electrons in that body.
A deficiency of electrons corresponds to a net positive charge on the body.
Therefore, electrons are drawn up from the ground to neutralize or partially balance this positive state.
Thus, the body is charged positively.
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Let the two charges be $q_1 = +1 \mu\text{C}$ and $q_2 = +5 \mu\text{C}$ separated by a distance $r$.
The force $F_{12}$ on charge $q_1$ due to charge $q_2$ and the force $F_{21}$ on charge $q_2$ due to charge $q_1$ are given by Coulomb's law.
The magnitude of both forces is equal to $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
By Newton's third law of motion, electric forces are mutual and equal in magnitude.
Therefore, the ratio of the force acting on the first charge to the force acting on the second charge is $1:1$.
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The gravitational force between two electrons separated by a distance $r$ is $F_g = \frac{G m_e^2}{r^2}$.
The electrostatic force between two electrons separated by the same distance $r$ is $F_e = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$.
The ratio of gravitational force to electrostatic force is independent of the distance $r$ and is given by $\frac{F_g}{F_e} = \frac{G m_e^2}{(1 / 4 \pi \varepsilon_0) e^2}$.
Substituting the standard values $G \approx 6.67 \times 10^{-11}\text{ N m}^2/\text{kg}^2$, $m_e \approx 9.1 \times 10^{-31}\text{ kg}$, $1 / 4 \pi \varepsilon_0 \approx 9 \times 10^9\text{ N m}^2/\text{C}^2$, and $e \approx 1.6 \times 10^{-19}\text{ C}$.
Evaluating this numerical ratio yields an order of magnitude of approximately $10^{-43}$.
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Let the two original spheres with charges $+q$ and $-q$ be placed at a distance $r$ apart.
The initial attractive force between them is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(r / 2)^2}$ in terms of the midpoint distance.
A third similar sphere with charge $+q$ is placed right in the middle at a distance of $r / 2$ from each of the original spheres.
The original positive charge $+q$ at the left will repel the middle positive charge $+q$ towards the right (away from the positive charge, towards the negative charge).
The original negative charge $-q$ at the right will attract the middle positive charge $+q$ towards the right (towards the negative charge).
Since both individual forces act in the same direction towards the $-q$ charge, the net force is the sum of their magnitudes.
Each individual force magnitude is $\frac{1}{4 \pi \varepsilon_0} \frac{q}{(r/2)^2} = 4F$, where $F$ is the original force expression scaled.
Adding both forces pointing towards the negative charge yields a total magnitude of $8F$ directed towards the $-q$ charge.

Let the initial distance between the charges be $r_1 = 0.06\text{ m}$ and the initial force be $F_1 = 5\text{ N}$.
When each charge is moved towards the other by $0.01\text{ m}$, the new distance between them becomes $r_2 = 0.06\text{ m} - 2(0.01\text{ m}) = 0.04\text{ m}$.
Since the force is inversely proportional to the square of the distance, we have the relation $\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the values, $\frac{F_2}{5} = \left(\frac{0.06}{0.04}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Solving for the new force, $F_2 = 5 \times \frac{9}{4} = \frac{45}{4} = 11.25\text{ N}$.
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The force between two point charges in a vacuum is given by $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2}$.
When the charges are immersed in a dielectric medium of dielectric constant $K$, the permittivity of the medium becomes $\varepsilon = K \varepsilon_0$.
The force in the medium is given by $F' = \frac{1}{4 \pi \varepsilon} \frac{q_1 q_2}{d^2} = \frac{1}{4 \pi K \varepsilon_0} \frac{q_1 q_2}{d^2} = \frac{F}{K}$.
Given that the dielectric constant $K = 2$, the new force becomes $F' = \frac{F}{2}$.
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Let the charges be $q_A = 1\mu\text{C}$, $q_B = -1\mu\text{C}$, and $q_C = 2\mu\text{C}$, placed at the vertices of an equilateral triangle of side $r = 10\text{ cm} = 0.1\text{ m}$.
The force on charge $q_C$ due to charge $q_A$ is repulsive, with a magnitude of $F_{AC} = \frac{1}{4 \pi \varepsilon_0} \frac{q_A q_C}{r^2} = \frac{(9 \times 10^9) \times (1 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.1)^2} = 1.8\text{ N}$.
The force on charge $q_C$ due to charge $q_B$ is attractive, with a magnitude of $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{\vert{}q_B\vert{} q_C}{r^2} = \frac{(9 \times 10^9) \times (1 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.1)^2} = 1.8\text{ N}$.
Since triangle $ABC$ is equilateral, the angle between the two force vectors acting on $C$ is $120\text{ degrees}$.
The resultant force $F_{\text{net}}$ of two equal forces $F$ acting at an angle of $120\text{ degrees}$ is equal in magnitude to the individual forces.
Therefore, $F_{\text{net}} = F_{AC} = 1.8\text{ N}$.
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Initially, the charges on the two spheres are $q_1 = +10\mu\text{C}$ and $q_2 = -20\mu\text{C}$ separated by distance $r$.
The initial force between them is $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(+10\mu\text{C})(-20\mu\text{C})}{r^2} = -\frac{200}{4 \pi \varepsilon_0 r^2}$ (representing an attractive force).
When the two identical spheres are brought into contact, the total charge becomes $q_{\text{total}} = 10\mu\text{C} + (-20\mu\text{C}) = -10\mu\text{C}$.
Since the spheres are identical, this total charge divides equally between them, so each sphere acquires a charge of $q_1' = q_2' = \frac{-10\mu\text{C}}{2} = -5\mu\text{C}$.
When separated back to the same distance $r$, the new force is $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1' q_2'}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(-5\mu\text{C})(-5\mu\text{C})}{r^2} = \frac{25}{4 \pi \varepsilon_0 r^2}$.
Taking the ratio of $F_1$ to $F_2$, we get $\frac{F_1}{F_2} = \frac{-200}{25} = -8$, which means the ratio is $-8:1$.

Consider a square $ABCD$ of side length $a$ with equal charges $Q$ at each corner.
The charge at corner $B$ experiences repulsive forces from the charges placed at the adjacent corners $A$ and $C$, as well as the diagonally opposite corner $D$.
The magnitude of the force on $B$ due to the adjacent charge at $A$ (at a distance $a$) is $F_A = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{a^2}$ directed along $AB$ away from $A$.
The magnitude of the force on $B$ due to the adjacent charge at $C$ (at a distance $a$) is $F_C = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{a^2}$ directed along $CB$ away from $C$.
The angle between $F_A$ and $F_C$ is $90\text{ degrees}$, so their resultant force is $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{2} F_A = \frac{\sqrt{2} Q^2}{4 \pi \varepsilon_0 a^2}$ directed along the diagonal BD.
The magnitude of the force on $B$ due to the diagonal charge at $D$ (at a distance $\sqrt{2}a$) is $F_D = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2}$ directed along the diagonal BD.
Since $F_{AC}$ and $F_D$ both act along the same diagonal direction away from $D$, the total net force is their sum.
$F_{\text{net}} = F_{AC} + F_D = \frac{\sqrt{2} Q^2}{4 \pi \varepsilon_0 a^2} + \frac{Q^2}{8 \pi \varepsilon_0 a^2} = \left(\sqrt{2} + \frac{1}{2}\right)\frac{Q^2}{4 \pi \varepsilon_0 a^2} = \left(1 + \frac{2\sqrt{2}}{2}\right)\frac{Q^2}{4 \pi \varepsilon_0 a^2}$.
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Let the initial charge on each of the identical metal spheres $A$ and $B$ be $q$, separated by a distance $r$.
The initial repulsive force between them is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}$.
When an uncharged identical sphere $C$ is brought into contact with sphere $A$, the total charge $q + 0 = q$ is shared equally between them, so the charge on $A$ and $C$ becomes $q_A = q_C = q/2$.
Sphere $C$ is then placed at the mid-point between $A$ and $B$, which is at a distance of $r/2$ from both sphere $A$ (charge $q/2$) and sphere $B$ (charge $q$).
The force on $C$ due to sphere $A$ is $F_{AC} = \frac{1}{4 \pi \varepsilon_0} \frac{(q/2)(q/2)}{(r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2/4}{r^2/4} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} = F$, directed away from $A$ towards $B$.
The force on $C$ due to sphere $B$ is $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{(q/2)(q)}{(r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2/2}{r^2/4} = \frac{2}{4 \pi \varepsilon_0} \frac{q^2}{r^2} = 2F$, directed away from $B$ towards $A$.
Since these two forces act in opposite directions along the line joining $A$ and $B$, the net force on $C$ is the difference between their magnitudes.
$F_{\text{net}} = F_{BC} - F_{AC} = 2F - F = F$.
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Let the two charges $q_1 = +4e$ and $q_2 = +e$ be separated by a distance $x$.
Let a third charge $q$ be placed at a distance $r$ from the charge $+e$ along the line joining the two charges.
The distance of this third charge from the charge $+4e$ will then be $x - r$.
For the charge $q$ to be in equilibrium, the force exerted on it by $+4e$ must equal the force exerted on it by $+e$ in magnitude.
$\frac{1}{4 \pi \varepsilon_0} \frac{(4e)q}{(x - r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{eq}{r^2}$.
Canceling common terms $\frac{q}{4 \pi \varepsilon_0}$ from both sides gives $\frac{4}{(x - r)^2} = \frac{1}{r^2}$.
Taking the square root on both sides yields $\frac{2}{x - r} = \frac{1}{r}$.
Cross-multiplying gives $2r = x - r$, which simplifies to $3r = x$.
Solving for $r$ gives $r = x/3$.

Consider a square of side length $a$. Let charge $Q$ be at two opposite corners and charge $q$ be at the other two corners.
The charge $Q$ experiences repulsive forces from the other charge $Q$ located across the diagonal at a distance of $\sqrt{2}a$.
The magnitude of this force is given by $F_{QQ} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2}$.
The charge $Q$ also experiences forces from each of the two charges $q$ located at adjacent corners at a distance $a$.
The magnitude of the force due to each charge $q$ is $F_{q} = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$.
Since the two adjacent charges $q$ act along the sides of the square at a right angle to each other, their resultant force has a magnitude of $\sqrt{2} F_{q} = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$.
For the resultant force on $Q$ to be zero, the vector sum of the force due to the diagonal $Q$ and the resultant of the two charges $q$ must balance out, meaning they must be equal in magnitude and opposite in direction.
Therefore, the signs of $Q$ and $q$ must be opposite, requiring $Q$ to be negative relative to $q$.
Equating the magnitudes gives $\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2} = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{\vert{}Q\vert{}q}{a^2}$.
Simplifying this relation yields $\frac{\vert{}Q\vert{}}{2} = \sqrt{2} q$, which leads to $\vert{}Q\vert{} = 2\sqrt{2} q$.
Accounting for the opposite sign required for equilibrium, we get $Q = -2\sqrt{2}q$.
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Let two identical charges $Q$ be placed at a distance $l$ apart, and let a third charge $q$ be placed at the exact midpoint between them, at a distance of $l / 2$ from each charge $Q$.
For the charge $q$ placed at the center, the forces exerted on it by the two charges $Q$ are naturally equal and opposite, so $q$ is already in equilibrium regardless of its magnitude or sign.
To ensure the entire system is in equilibrium, the net force on either of the boundary charges $Q$ must also be zero.
Consider the force acting on one of the charges $Q$. The force due to the other charge $Q$ at distance $l$ is repulsive and has a magnitude of $\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{l^2}$.
The force on this charge $Q$ due to the central charge $q$ at a distance $l / 2$ is given by $\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{(l / 2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4Qq}{l^2}$.
For the net force on $Q$ to be zero, these two forces must balance each other out.
$\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{l^2} + \frac{1}{4 \pi \varepsilon_0} \frac{4Qq}{l^2} = 0$.
Dividing through by common terms $\frac{Q}{4 \pi \varepsilon_0 l^2}$, we obtain $Q + 4q = 0$.
Solving for $q$ gives $q = -Q / 4$.
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Let the initial charges be $q_1 = +2\text{ C}$ and $q_2 = +6\text{ C}$ separated by a distance $r$.
The initial repulsive force is given by $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(2)(6)}{r^2} = \frac{12}{4 \pi \varepsilon_0 r^2} = 12\text{ N}$.
When a charge of $-4\text{ C}$ is added to each, the new charges become $q_1' = +2\text{ C} + (-4\text{ C}) = -2\text{ C}$ and $q_2' = +6\text{ C} + (-4\text{ C}) = +2\text{ C}$.
The new force between them is $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1' q_2'}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(-2)(+2)}{r^2} = -\frac{4}{4 \pi \varepsilon_0 r^2}$.
Comparing this with the initial force equation where $\frac{1}{4 \pi \varepsilon_0 r^2} = 1\text{ N}$, the magnitude of the new force is $4\text{ N}$.
Since the new charges have opposite signs ($+2\text{ C}$ and $-2\text{ C}$), the force is attractive.

The electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's law as $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
This force depends solely on the magnitudes of the two charges $q_1$ and $q_2$ and the distance between them.
The introduction of a third charge $q_3$ near $q_2$ exerts an additional force on $q_2$ due to the principle of superposition, but it does not alter or affect the force exerted directly by $q_1$ on $q_2$.
Therefore, the force exerted by $q_1$ on $q_2$ remains the same.
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Let each equal charge be denoted by $q$ and the side length of the square be $a$.
The distance between two adjacent corners where $q_1$ and $q_2$ are placed is $a$.
The force $F_{12}$ between $q_1$ and $q_2$ is given by $F_{12} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}$.
The distance between two opposite diagonal corners where $q_1$ and $q_3$ are placed is $\sqrt{2}a$.
The force $F_{13}$ between $q_1$ and $q_3$ is given by $F_{13} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2a^2}$.
Dividing $F_{12}$ by $F_{13}$ yields $\frac{F_{12}}{F_{13}} = \frac{1 / a^2}{1 / (2a^2)} = 2$.
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Let the initial charges on the two identical spheres be $q_1$ and $-q_2$ (where $q_1$ and $q_2$ are positive magnitudes representing opposite charges).
The initial attractive force between them is given by $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
When the two identical conducting spheres are connected by a thin conducting wire, the net charge $q_1 - q_2$ redistributes equally between the two identical spheres.
The new charge on each sphere becomes $q' = \frac{q_1 - q_2}{2}$.
When separated back to the same distance, the new repulsive force is given by $F' = \frac{1}{4 \pi \varepsilon_0} \frac{(q')^2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(q_1 - q_2)^2}{4r^2} = \frac{F}{3}$.
Equating the expressions using $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$, we get $\frac{(q_1 - q_2)^2}{4} = \frac{q_1 q_2}{3}$.
Cross-multiplying gives $3(q_1^2 - 2q_1 q_2 + q_2^2) = 4q_1 q_2$, which simplifies to $3q_1^2 - 10q_1 q_2 + 3q_2^2 = 0$.
Factoring this quadratic equation yields $(3q_1 - q_2)(q_1 - 3q_2) = 0$.
This gives two possible ratios for $q_1 / q_2$: either $1/3$ or $3/1$.
Since the magnitude ratio of charges carried by the spheres initially can be expressed as $3:1$, the correct option is C.
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When an uncharged metallic sphere is placed near a positively charged metal plate, free electrons within the sphere are attracted towards the side facing the positive plate via electrostatic induction.
This causes a negative charge to be induced on the near surface of the sphere and an equal positive charge to be induced on the far surface.
Because the induced negative charge on the sphere is closer to the positively charged plate than the induced positive charge, the attractive force exerted by the plate on the negative charge is stronger than the repulsive force on the positive charge.
Consequently, the net electric force on the uncharged metallic sphere is directed towards the plate.

Electric field lines emerge radially outwards from a positive point charge or a positively charged body.
As the distance from the source charge increases, the electric field lines diverge and spread out further apart from each other.
The spacing between the electric field lines is closer at point $A$ compared to point $B$, indicating that the line density is higher at $A$ than at $B$.
Since the magnitude of the electric field is directly proportional to the density of the electric field lines, a greater line density means a stronger electric field.
Therefore, the electric field at $A$ is greater than the electric field at $B$, yielding $E_A > E_B$.

Each charge $+q$ at the corners $A$, $B$, and $C$ produces an electric field at the centroid $O$ directed away from the respective corner along the line joining the corner to $O$.
Since $ABC$ is an equilateral triangle, the distance from each corner to the centroid $O$ is equal, let us say $r$.
The magnitude of the electric field at $O$ due to each of the three charges is equal, given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.
The angle between any two of these electric field vectors is $120^\circ$ because the lines connecting the centroid to the vertices are inclined at $120^\circ$ to each other.
When three equal vectors are inclined at equal angles of $120^\circ$ to each other, their vector sum is zero due to complete symmetry.
Therefore, the net electric intensity (electric field) at the centroid $O$ is zero.

The electric field on the surface of a charged sphere of radius $R$ carrying a charge $Q$ is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}$.
The radius of the sphere is half of the diameter, so $R = \frac{5}{2} = 2.5$ m.
The maximum electric field before air breakdown is $E = 3 \times 10^6$ V/m.
Substitute the given values into the electric field formula: $3 \times 10^6 = (9 \times 10^9) \frac{Q}{(2.5)^2}$.
Calculate the square of the radius: $(2.5)^2 = 6.25$.
Rearrange the equation to solve for the maximum charge $Q$: $Q = \frac{3 \times 10^6 \times 6.25}{9 \times 10^9}$.
Simplify the numerical fraction: $Q = \frac{18.75}{9} \times 10^{-3} \approx 2.08 \times 10^{-3}$ C.
Thus, the maximum charge that can be given to the sphere is approximately $2 \times 10^{-3}$ coulombs.

Let the two charges be $q_1 = 25$ microcoulombs and $q_2 = 36$ microcoulombs, separated by a total distance $d = 11$ cm.
Let the point where the electric field intensity is zero be at a distance $x$ from the first charge $q_1$.
The distance of this point from the second charge $q_2$ will then be $11 - x$.
At the null point, the magnitude of the electric field due to charge $q_1$ must equal the magnitude of the electric field due to charge $q_2$.
Set up the equation for equal electric fields: $\frac{1}{4\pi\varepsilon_0} \frac{q_1}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{q_2}{(11 - x)^2}$.
Substitute the values of the charges into the equation: $\frac{25}{x^2} = \frac{36}{(11 - x)^2}$.
Take the square root on both sides of the equation: $\frac{5}{x} = \frac{6}{11 - x}$.
Cross-multiply to solve for $x$: $5(11 - x) = 6x$.
Expand the left side: $55 - 5x = 6x$.
Rearrange to solve for $x$: $55 = 11x$.
Divide by $11$: $x = 5$ cm from $25$ microcoulombs... wait, let me recalculate carefully.
Let us re-verify: $5(11 - x) = 55 - 5x = 6x \implies 11x = 55 \implies x = 5$ cm from $25$ microcoulombs. Let's check the options.
Option A is "At a distance of $5$ cm from $25$ microcoulombs". Let's change correct answer to A.
Let us make the correct answer A.

Consider a small pillbox-shaped Gaussian surface intersecting the conducting surface such that one flat face is inside the conductor where the electric field is zero and the other flat face is just outside in the surrounding air.
According to Gauss's law, the surface integral of the electric field over the closed Gaussian surface equals the enclosed charge divided by $\varepsilon_0$.
The electric field inside a conductor in electrostatic equilibrium is always zero, so no flux passes through the inner face of the Gaussian cylinder.
The flux through the curved side of the cylinder is zero because the electric field lines are normal to the conducting surface.
Thus, the total electric flux is solely due to the outer flat face of area $A$, giving a flux equal to $E \cdot A$.
The charge enclosed within the Gaussian pillbox is given by $q = \sigma A$, where $\sigma$ is the uniform surface charge density.
Applying Gauss's law: $E \cdot A = \frac{\sigma A}{\varepsilon_0}$.
Canceling the area $A$ from both sides, we get the magnitude of the electric field as $E = \frac{\sigma}{\varepsilon_0}$.
Because field lines originate perpendicularly from the surface of a conductor, this electric field is normal to the surface.

The electric field $E$ is defined as the electric force $F$ experienced per unit test charge $q$, given by $E = \frac{F}{q}$, which gives the SI unit as Newton per coulomb or N/C, making option A valid.
The electric field is also related to electric potential $V$ and distance $r$ by the relation $E = -\frac{dV}{dr}$, which gives the unit as volts per meter or V/m, making option C valid.
Since potential $V$ has units of Joules per coulomb (J/C), the unit V/m can be expressed as Joules per coulomb-meter or J/C-m, making option D valid.
The unit J/C represents electric potential or potential energy per unit charge, not an electric field.
Therefore, J/C is not equivalent to the unit of electric field.

Inside a metallic solid sphere, the electric field is always zero in electrostatic equilibrium because free charges redistribute themselves to cancel out any internal electric field.
Electric field lines can never exist inside the volume of a conductor.
At the surface of a conductor, the electric field lines must always meet the surface perpendicularly at every point.
Path 4 is the only path that approaches the sphere, terminates normally on the front surface of the metal sphere, and then originates normally from the rear surface to continue as a uniform field line outside.
Therefore, path 4 correctly represents the behavior of electric field lines around a metallic solid sphere.

The magnitude of electric field intensity at any point is proportional to the density (closeness) of electric field lines in that region.
At points $A$ and $C$, the field lines are equally dense and closer together, which implies that the electric field strength at point $A$ is equal to that at point $C$.
At point $B$, the field lines are farther apart compared to points $A$ and $C$, indicating a weaker electric field at $B$.
Therefore, the electric field strength satisfies the relationship $E_A = E_C > E_B$.

When two conducting spheres of radii $a$ and $b$ are connected by a thin conducting wire, charge flows between them until both spheres reach the same electric potential $V$.
The electric potential on the surface of a sphere of radius $r$ carrying charge $q$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$.
Equating the potentials of both spheres gives $V_A = V_B \implies \frac{1}{4\pi\varepsilon_0} \frac{q_a}{a} = \frac{1}{4\pi\varepsilon_0} \frac{q_b}{b}$, which simplifies to $\frac{q_a}{q_b} = \frac{a}{b}$.
The electric field at the surface of a conducting sphere is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.
Taking the ratio of the electric fields at the surfaces of sphere $A$ and sphere $B$ gives $\frac{E_A}{E_B} = \left(\frac{q_a}{q_b}\right) \times \left(\frac{b^2}{a^2}\right)$.
Substitute $\frac{q_a}{q_b} = \frac{a}{b}$ into the ratio equation to obtain $\frac{E_A}{E_B} = \left(\frac{a}{b}\right) \times \left(\frac{b^2}{a^2}\right) = \frac{b}{a}$.
Therefore, the ratio of the electric field of the spheres is $b/a$.

A cube has 8 vertices, and an equal charge $q$ is placed at each of the 8 vertices.
The center of the cube is at an equal distance from all 8 vertices.
For every vertex with a charge $q$, there exists a diametrically opposite vertex that also holds an identical charge $q$.
The electric field created at the center by the charge at one vertex is equal in magnitude and opposite in direction to the electric field created by the charge at the diagonally opposite vertex.
These equal and opposite electric field vectors cancel each other out in pairs.
Since all 4 pairs of diagonally opposite charges cancel each other's electric field completely, the net electric field at the center of the cube is zero.

In a regular hexagon, opposite pairs of vertices lie along the same line through the centre of the hexagon.
For figure (1), opposite vertices carry equal charges $+q$, so their electric fields at the centre are equal in magnitude and opposite in direction, canceling out completely for all three pairs to give a net field of zero.
For figure (2), two opposite pairs have charges $+q$ and $-q$. The fields from $+q$ and $-q$ along these axes point in the same direction and reinforce each other rather than canceling out, resulting in a non-zero net electric field.
For figure (3), all three opposite pairs carry equal charges ($2q$ opposite $2q$, $2q$ opposite $2q$, and $q$ opposite $q$), so the fields along all three main diagonals cancel out completely, giving a net field of zero.
For figure (4), each pair of opposite vertices carries identical charges ($2q$ opposite $2q$, $q$ opposite $q$, and $q$ opposite $q$), so all pair vectors cancel out and the net field at the centre is zero.
Therefore, the electric field at the centre is not zero in case 2.