Mix Examples-Electric Charges and Fields Questions in English

(C) Statement-$I$ is true: When a positive point charge is placed in an existing electric field,the total electric field at any point is the vector sum of the external field and the field produced by the point charge. Depending on the position relative to the charge,the field can increase or decrease.
Statement-$II$ is false: While it is generally true that a dipole in a non-uniform field experiences a net force,it is not always true. If the dipole is placed at a point where the electric field has a local extremum (maximum or minimum),the net force on the dipole can be zero. Therefore,the statement that the force will 'not be zero' is not universally true.

(A) The electric displacement vector is defined as $\overrightarrow{D} = \epsilon_{0} \overrightarrow{E}$.
Dimensional analysis of electric displacement: $[D] = [\epsilon_{0}][E]$.
Since the electric field $E$ is related to surface charge density $\sigma$ by $E = \frac{\sigma}{\epsilon_{0}}$,we have $\epsilon_{0} E = \sigma$.
Therefore,the dimensions of electric displacement $[D]$ are equal to the dimensions of surface charge density $[\sigma]$.
Both have dimensions of $[Q L^{-2}]$,where $Q$ is charge and $L$ is length.

(B) Let the initial charges on spheres $A$ and $B$ be $q_A = q_B = q$. The initial force is $F = \frac{Kq^2}{r^2}$.
When sphere $C$ (initially uncharged) is placed in contact with $A$,the charge redistributes equally: $q_A' = q_C' = \frac{q}{2}$.
Now,sphere $C$ is placed in contact with $B$. The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. Upon separation,each gets $q_B' = q_C'' = \frac{3q/2}{2} = \frac{3q}{4}$.
Now,$A$ has charge $\frac{q}{2}$ and $B$ has charge $\frac{3q}{4}$. Sphere $C$ is placed at the midpoint (distance $r/2$ from both).
The force on $C$ due to $A$ is $F_1 = \frac{K(q/2)(3q/4)}{(r/2)^2} = \frac{3Kq^2/8}{r^2/4} = \frac{3Kq^2}{2r^2} = \frac{3F}{2}$ (towards $A$).
The force on $C$ due to $B$ is $F_2 = \frac{K(3q/4)(3q/4)}{(r/2)^2} = \frac{9Kq^2/16}{r^2/4} = \frac{9Kq^2}{4r^2} = \frac{9F}{4}$ (towards $A$).
The net force on $C$ is $F_{net} = F_2 - F_1 = \frac{9F}{4} - \frac{3F}{2} = \frac{9F - 6F}{4} = \frac{3F}{4}$.

(C) For an uncharged particle in projectile motion,the range is $L = \frac{u^2 \sin 2\theta}{g} \quad \dots(i)$
For a charged particle,the effective acceleration becomes $g' = g + \frac{qE}{m}$.
Given that the range for a particle with charge $q$ is $L/2$,we have:
$\frac{L}{2} = \frac{u^2 \sin 2\theta}{g + \frac{qE}{m}}$
Substituting $u^2 \sin 2\theta = Lg$ from equation $(i)$:
$\frac{L}{2} = \frac{Lg}{g + \frac{qE}{m}}$
$\Rightarrow g + \frac{qE}{m} = 2g \Rightarrow \frac{qE}{m} = g \quad \dots(ii)$
Now,for a particle of mass $m$ and charge $2q$,the effective acceleration is $g'' = g + \frac{2qE}{m}$.
Using equation $(ii)$,$g'' = g + 2g = 3g$.
The new range $R$ is:
$R = \frac{u^2 \sin 2\theta}{3g} = \frac{1}{3} \left( \frac{u^2 \sin 2\theta}{g} \right) = \frac{L}{3}$.

(C) Using the method of images for a grounded conducting corner,we place three image charges to satisfy the boundary conditions: $-q$ at $(-d, 0)$,$-q$ at $(0, -d)$,and $+q$ at $(-d, -d)$.
The force on the charge $+q$ at $(d, d)$ due to the image charges is:
$1$. Force due to $-q$ at $(-d, 0)$: $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(2d)^2 + 0^2} = \frac{q^2}{16 \pi \varepsilon_0 d^2}$ (attractive,towards the $x$-axis).
$2$. Force due to $-q$ at $(0, -d)$: $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{0^2 + (2d)^2} = \frac{q^2}{16 \pi \varepsilon_0 d^2}$ (attractive,towards the $y$-axis).
$3$. Force due to $+q$ at $(-d, -d)$: $F_3 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(2d)^2 + (2d)^2} = \frac{q^2}{32 \pi \varepsilon_0 d^2}$ (repulsive,away from the origin).
The resultant attractive force $F_{12}$ from the two $-q$ charges is $\sqrt{F_1^2 + F_2^2} = \sqrt{2} F_1 = \frac{\sqrt{2} q^2}{16 \pi \varepsilon_0 d^2}$,directed towards the origin $O$.
The net force $F_{\text{net}}$ is $F_{12} - F_3 = \frac{\sqrt{2} q^2}{16 \pi \varepsilon_0 d^2} - \frac{q^2}{32 \pi \varepsilon_0 d^2} = \frac{q^2}{32 \pi \varepsilon_0 d^2} (2\sqrt{2} - 1)$,directed towards $O$.

(C) Let $q_1$ and $q_2$ be the charges at two vertices of an equilateral triangle of side $a$. The magnitude of the electric field $E$ at the third vertex is given by the vector sum of fields $E_1$ and $E_2$ due to $q_1$ and $q_2$ respectively.
$E = \sqrt{E_1^2 + E_2^2 + 2 E_1 E_2 \cos 60^{\circ}}$
Since $E_1 = \frac{k q_1}{a^2}$ and $E_2 = \frac{k q_2}{a^2}$,we have:
$E = \frac{k}{a^2} \sqrt{q_1^2 + q_2^2 + q_1 q_2}$
Given $q_1 + q_2 = q$,let $x = q_1 / q$,so $q_1 = xq$ and $q_2 = (1-x)q$.
Substituting these into the expression for $E$:
$E = \frac{k}{a^2} \sqrt{(xq)^2 + ((1-x)q)^2 + (xq)((1-x)q)}$
$E = \frac{kq}{a^2} \sqrt{x^2 + 1 - 2x + x^2 + x - x^2} = \frac{kq}{a^2} \sqrt{x^2 - x + 1}$
To find the minimum,we differentiate $f(x) = x^2 - x + 1$ with respect to $x$ and set it to zero:
$f'(x) = 2x - 1 = 0 \implies x = 0.5$.
At $x = 0$ or $x = 1$,$E = \frac{kq}{a^2}$. At $x = 0.5$,$E = \frac{kq}{a^2} \sqrt{0.25 - 0.5 + 1} = \frac{kq}{a^2} \sqrt{0.75} = \frac{\sqrt{3}}{2} \frac{kq}{a^2}$.
This corresponds to the curve in graph $C$.

(C) $1$. Repulsion occurs only between like charges. Thus,$b$ and $c$ have the same charge,and $d$ and $e$ have the same charge.
$2$. Attraction occurs between opposite charges or between a charged body and an uncharged body (due to induction).
$3$. Given $(a, b)$ attract,$(c, e)$ attract,and $(a, e)$ attract.
$4$. If $a$ were charged (say positive),then $b$ must be negative. Since $(b, c)$ repel,$c$ must also be negative. Since $(c, e)$ attract,$e$ must be positive (or uncharged). If $e$ is positive,then $(a, e)$ would repel,which contradicts the given information. If $e$ is uncharged,then $(a, e)$ would attract,but $(d, e)$ would not repel (as $d$ would need to be charged to repel $e$).
$5$. If $a$ is uncharged,it will attract any charged body $b$ or $e$ due to electrostatic induction. This is consistent with all given observations.

(D) The principle of electrostatic induction is a fundamental phenomenon used in various industrial and natural processes.
$1$. Pollination: In many plants,pollen grains are attracted to the stigma due to electrostatic forces generated by induction.
$2$. Chocolate making: Electrostatic deposition is used to coat chocolates uniformly.
$3$. Xerox copying: The process of xerography (photocopying) relies heavily on electrostatic induction to transfer toner particles onto paper.
Since all these processes utilize the principle of electrostatic induction,the correct option is $D$.

(C) For a regular hexagon with side $a$,the distance from each vertex to the center $O$ is $r = a$.
$1$. Electric Potential $(V)$: The potential at the center $O$ is the algebraic sum of potentials due to individual charges: $V = \sum \frac{k q_i}{r_i} = \frac{k}{a} (q - q + q - q + q - q) = 0$.
$2$. Electric Field $(E)$: The electric field at the center $O$ is the vector sum of fields due to individual charges. Each pair of opposite charges (e.g.,$q$ and $-q$) produces a resultant field at the center directed towards the negative charge. Specifically,a charge $q$ at one vertex and $-q$ at the opposite vertex create a net field of magnitude $\frac{2kq}{a^2}$ directed towards the negative charge. Since there are three such pairs oriented at $120^\circ$ to each other,their vector sum is zero. Thus,$E = 0$.

(A) The electric field due to an infinite plane sheet of charge is given by $E_S = \frac{|\sigma_s|}{2\epsilon_0}$.
At the point $(0, 0, 2)$,the distance from the sheet (in the $x-y$ plane,i.e.,$z=0$) is $r_S = 2 \text{ m}$.
The electric field due to an infinitely long line charge is given by $E_{\ell} = \frac{|\lambda_e|}{2\pi\epsilon_0 r_{\ell}}$.
The line charge is at $z=4 \text{ m}$ and is parallel to the $y$-axis. The point is $(0, 0, 2)$. The perpendicular distance $r_{\ell}$ from the line charge to the point $(0, 0, 2)$ is $|4 - 2| = 2 \text{ m}$.
Given $|\sigma_s| = 2|\lambda_e|$.
The ratio of the electric fields is:
$\frac{E_S}{E_{\ell}} = \frac{|\sigma_s| / 2\epsilon_0}{|\lambda_e| / 2\pi\epsilon_0 r_{\ell}} = \frac{|\sigma_s|}{2\epsilon_0} \times \frac{2\pi\epsilon_0 r_{\ell}}{|\lambda_e|} = \frac{|\sigma_s| \pi r_{\ell}}{|\lambda_e|}$.
Substituting the values $|\sigma_s| = 2|\lambda_e|$ and $r_{\ell} = 2 \text{ m}$:
$\frac{E_S}{E_{\ell}} = \frac{2|\lambda_e| \times \pi \times 2}{|\lambda_e|} = 4\pi$.
We are given the ratio as $\pi \sqrt{n} : 1$,so $\pi \sqrt{n} = 4\pi$,which implies $\sqrt{n} = 4$.
Therefore,$n = 16$.

(C) The charges are $q_A = q/3$,$q_B = q/3$,and $q_C = -2q/3$.
In $\triangle ABC$,since $A, B, C$ are on a circle with center $O$,$OA = OB = OC = R$.
Given $\angle CAB = 60^{\circ}$,in $\triangle OAC$,$OA = OC = R$,so $\angle OCA = \angle OAC = 60^{\circ}$,implying $\triangle OAC$ is equilateral. Thus,$AC = R$.
Similarly,for $\triangle OBC$,since $\angle ACB = 90^{\circ}$ (angle in a semicircle if $AB$ is diameter,but here $AB$ is a chord),we calculate the distance $BC$.
Using the law of cosines in $\triangle OBC$,or geometry: $BC = \sqrt{R^2 + R^2 - 2R^2 \cos(120^{\circ})} = R\sqrt{3}$.
The force between $C$ and $B$ is $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_C| |q_B|}{(BC)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(2q/3)(q/3)}{(R\sqrt{3})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2q^2/9}{3R^2} = \frac{q^2}{54 \pi \varepsilon_0 R^2}$.
Thus,option $C$ is correct.

(A) The correct option is $A$.
In a central force field,such as the Coulomb field created by a stationary charge $+Q$,the force acting on the orbiting charge $-q$ is always directed towards the center (the position of $+Q$).
Since the force is a central force,the torque $\vec{\tau} = \vec{r} \times \vec{F}$ acting on the charge $-q$ about the center $+Q$ is zero.
According to the principle of conservation of angular momentum,if the net external torque on a system is zero,the angular momentum $\vec{L}$ remains constant.
Therefore,the angular momentum of the charge $-q$ is constant throughout its elliptical orbit.
Linear momentum,angular velocity,and linear speed are not constant in an elliptical orbit because the distance between the charges changes,leading to variations in speed and direction.

(A) The electrostatic pressure on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\varepsilon_0}$.
This pressure acts radially outward on every element of the surface.
To keep the two hemispherical shells together, the external force $F$ must balance the total outward force exerted by the electrostatic pressure on the cross-sectional area of the hemisphere.
The cross-sectional area of the hemisphere is $A = \pi R^2$.
The total outward force $F$ is the product of the electrostatic pressure and the projected area perpendicular to the force direction:
$F = P \times A = \left( \frac{\sigma^2}{2\varepsilon_0} \right) \times \pi R^2$.
Thus, $F = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$.
Since $\pi$, $2$, and $\varepsilon_0$ are constants, the force $F$ is proportional to $\sigma^2 R^2$.

(D) $1$. When the drop is balanced: $qE = mg = \frac{4}{3} \pi R^3 \rho g$ (assuming air density is negligible).
$2$. When the field is switched off,the drop falls with terminal velocity $v_T$: $mg = 6 \pi \eta R v_T$.
$3$. From the second equation,$R = \sqrt{\frac{9 \eta v_T}{2 \rho g}}$.
$4$. Substituting values: $R = \sqrt{\frac{9 \times 1.8 \times 10^{-5} \times 2 \times 10^{-3}}{2 \times 900 \times 9.8}} = \sqrt{\frac{32.4 \times 10^{-8}}{17640}} \approx 1.355 \times 10^{-6} \text{ m}$.
$5$. Now,$q = \frac{6 \pi \eta R v_T}{E} = \frac{6 \pi \times 1.8 \times 10^{-5} \times 1.355 \times 10^{-6} \times 2 \times 10^{-3}}{\frac{81 \pi}{7} \times 10^5}$.
$6$. Calculating this yields $q = 8.0 \times 10^{-19} \text{ C}$.

(A) Consider the equilibrium of one side of the square film,say side $BC$ of length $a$. The surface tension force acting on this side is $F_2 = \gamma a$ (since a soap film has two surfaces,the force is $2 \gamma a$,but here we consider the force per unit length acting on the film edge).
The electrostatic force on the charges at $B$ and $C$ due to the other charges must be balanced by the surface tension. The net electrostatic force on charge $B$ due to charges at $A$ and $C$ is $F_{AC} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2} \sqrt{2}$ (along the diagonal $DB$) and due to charge $D$ is $F_D = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2a^2}$ (along $DB$).
The component of the net electrostatic force on the side $BC$ perpendicular to $BC$ is $F_{net} = 2 \times F_{charge} \cos(45^{\circ})$.
Equating the forces: $2 \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2} \right) \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \gamma a$.
Simplifying,$a^3 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{\gamma} \left( \sqrt{2} + \frac{1}{2} \right)$.
Comparing this with $a = k \left[ \frac{q^2}{\gamma} \right]^{1/N}$,we get $N = 3$.

(C) The electric field lines originate from a positive charge and terminate at a negative charge. They do not form closed loops.
Magnetic field lines,on the other hand,always form closed continuous loops because magnetic monopoles do not exist.
However,the question asks for a pattern valid for both.
Option $C$ shows circular field lines.
For an electric field,circular lines can exist in regions with time-varying magnetic fields (induced electric fields).
For a magnetic field,circular lines are produced by a straight current-carrying wire.
Thus,the circular field pattern is the only one that can represent both an electric field (in specific non-electrostatic conditions) and a magnetic field.

(B) The correct statements are $(C)$ and $(D)$.
$(A)$ Gauss's law is derived from the inverse-square law $(E \propto r^{-2})$. If the field varies as $r^{-2.5}$,the flux through a closed surface would depend on the shape and size of the surface,making the standard Gauss law invalid.
$(B)$ Gauss's law is most effective when there is high symmetry (spherical,cylindrical,or planar). An electric dipole lacks the symmetry required to simplify the calculation of the electric field using Gauss's law.
$(C)$ For two point charges,the electric field is zero at a point between them only if the charges have the same sign (repulsive force). If they have opposite signs,the field is zero outside the region between them.
$(D)$ By definition,the potential difference $V_B - V_A$ is the work done by an external agent in moving a unit positive charge from $A$ to $B$ without acceleration.

(B) $(1)$ Electric field due to a point charge at the origin is $E = \frac{kQ}{d^2}$,so $E \propto \frac{1}{d^2}$.
$(2)$ Electric field at any point on the axis of a dipole is $E = \frac{2kp}{d^3} = \frac{4kQl}{d^3}$,so $E \propto \frac{1}{d^3}$.
$(3)$ Electric field due to an infinite long line charge is $E = \frac{2k\lambda}{d}$,so $E \propto \frac{1}{d}$.
$(4)$ Electric field due to two infinite long wires is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{2k\lambda}{d-l} - \frac{2k\lambda}{d+l} = \frac{4k\lambda l}{d^2-l^2}$. If $d \gg l$,then $E = \frac{4k\lambda l}{d^2}$,so $E \propto \frac{1}{d^2}$.
$(5)$ Electric field due to an infinite plane charge is $E = \frac{\sigma}{2\epsilon_0}$,which is independent of $d$.
Matching: $P \rightarrow 5$,$Q \rightarrow 3$,$R \rightarrow 1, 4$,$S \rightarrow 2$. Thus,option $B$ is correct.

(A) Let the initial charge on the sphere be $Q$. Thus,$V_0 = \frac{kQ}{R}$.
When a small hole of area $\alpha(4\pi R^2)$ is made,the charge removed is $q = \alpha Q$.
$(1)$ Potential at the center $(V_c)$ and at a point $P$ at distance $R/2$ from the center towards the hole $(V_p)$:
The potential at any point is the sum of the potential due to the complete sphere and the potential due to the removed charge element (treated as a negative point charge at the surface).
$V_c = \frac{kQ}{R} - \frac{kq}{R} = \frac{kQ}{R}(1-\alpha) = V_0(1-\alpha)$.
$V_p = \frac{kQ}{R} - \frac{kq}{R/2} = \frac{kQ}{R} - \frac{2kq}{R} = \frac{kQ}{R}(1-2\alpha) = V_0(1-2\alpha)$.
Therefore,the ratio $\frac{V_c}{V_p} = \frac{1-\alpha}{1-2\alpha}$. Thus,option $(A)$ is correct.
$(2)$ Electric field at the center $(E_c)$:
Initially,$E_c = 0$. After removing charge $q$,$E_c = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \alpha \frac{V_0}{R}$. The field increases.
$(3)$ Electric field at a point $P'$ at distance $2R$ from the center:
Initially,$E_{P'} = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2}$.
After removing charge $q$ from the surface,the field due to the hole at $P'$ is $\frac{kq}{R^2}$ (since $P'$ is at distance $R$ from the hole).
$E_{P', \text{final}} = \frac{kQ}{4R^2} - \frac{kq}{R^2} = \frac{kQ}{4R^2} - \frac{k(\alpha Q)}{R^2} = \frac{kQ}{4R^2} - \frac{4k\alpha Q}{4R^2} = \frac{kQ}{4R^2}(1-4\alpha)$.
Change in field $\Delta E = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \frac{\alpha V_0}{R}$.

(C) The electric fields are given by: $E_1(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$,$E_2(r) = \frac{\lambda}{2 \pi \epsilon_0 r}$,and $E_3(r) = \frac{\sigma}{2 \epsilon_0}$.
Given $E_1(r_0) = E_2(r_0) = E_3(r_0) = E_0$,we have:
$\frac{Q}{4 \pi \epsilon_0 r_0^2} = \frac{\lambda}{2 \pi \epsilon_0 r_0} = \frac{\sigma}{2 \epsilon_0} = E_0$.
From $E_2(r_0) = E_3(r_0)$,we get $\frac{\lambda}{2 \pi \epsilon_0 r_0} = \frac{\sigma}{2 \epsilon_0}$,which implies $r_0 = \frac{\lambda}{\pi \sigma}$. Thus,option $B$ is incorrect.
From $E_1(r_0) = E_3(r_0)$,we get $\frac{Q}{4 \pi \epsilon_0 r_0^2} = \frac{\sigma}{2 \epsilon_0}$,which implies $Q = 2 \pi \sigma r_0^2$. Thus,option $A$ is incorrect.
Now,check option $C$: $E_1(r_0/2) = \frac{Q}{4 \pi \epsilon_0 (r_0/2)^2} = 4 E_1(r_0) = 4 E_0$. Also,$E_2(r_0/2) = \frac{\lambda}{2 \pi \epsilon_0 (r_0/2)} = 2 E_2(r_0) = 2 E_0$. Therefore,$E_1(r_0/2) = 2 E_2(r_0/2)$. Option $C$ is correct.
Check option $D$: $E_3(r)$ is independent of $r$,so $E_3(r_0/2) = E_3(r_0) = E_0$. Since $E_2(r_0/2) = 2 E_0$,$E_2(r_0/2) = 2 E_3(r_0/2)$. Option $D$ is incorrect.

(A) Let the force exerted by charge $Q_i$ on $q$ be $\vec{F}_i$. The $x$-component of the force from a pair of charges at $\pm x_0$ is proportional to $(Q_{left} - Q_{right})$. The $y$-component is proportional to $(Q_{left} + Q_{right})$.
$(P)$ All positive: $Q_1=Q_2=Q_3=Q_4 = +Q$. The $x$-components cancel out due to symmetry. The $y$-components add up, resulting in a net force along $+y$ (Option $3$).
$(Q)$ $Q_1, Q_2$ positive, $Q_3, Q_4$ negative: The $y$-components cancel out. The $x$-components add up in the $+x$ direction because the positive charges are on the left and negative on the right, pushing $q$ to the right (Option $1$).
$(R)$ $Q_1, Q_4$ positive, $Q_2, Q_3$ negative: The $x$-components cancel out. The $y$-components are negative because the charges closer to the origin $(Q_2, Q_3)$ are negative and stronger in their $y$-effect than the farther positive charges $(Q_1, Q_4)$, resulting in a net force along $-y$ (Option $4$).
$(S)$ $Q_1, Q_3$ positive, $Q_2, Q_4$ negative: The $y$-components cancel out. The $x$-components result in a net force along $-x$ because the net effect of the charges on the left is negative and on the right is positive, pulling $q$ to the left (Option $2$).
Thus, the correct matching is $P-3, Q-1, R-4, S-2$.

(A) The potential energy of the particle is $U(z) = qV(z) = \frac{q\sigma}{2\epsilon_0}(\sqrt{R^2+z^2} - z)$.
The external force is $\vec{F} = -c\hat{k}$,so the potential energy due to this force is $U_{ext}(z) = cz$.
The total potential energy is $U_{total}(z) = \frac{q\sigma}{2\epsilon_0}(\sqrt{R^2+z^2} - z) + cz$.
Using $\beta = \frac{2c\epsilon_0}{q\sigma}$,we have $c = \frac{\beta q\sigma}{2\epsilon_0}$.
Thus,$U_{total}(z) = \frac{q\sigma}{2\epsilon_0}(\sqrt{R^2+z^2} - z + \beta z)$.
For the particle to reach the origin,the total energy at $z_0$ must be greater than or equal to the potential energy at $z=0$. Since the particle starts from rest,$E = U_{total}(z_0)$.
$U_{total}(z_0) \ge U_{total}(0) \implies \sqrt{R^2+z_0^2} - z_0 + \beta z_0 \ge R$.
For $(A)$: $\beta = 1/4, z_0 = 25/7 R$. $\sqrt{R^2 + (25/7)^2 R^2} - 25/7 R + 1/4(25/7) R = \sqrt{1 + 625/49} R - 25/7 R + 25/28 R = \sqrt{674/49} R - 75/28 R \approx 3.71 R - 2.67 R = 1.04 R > R$. Particle reaches origin.
For $(B)$: $\beta = 1/4, z_0 = 3/7 R$. $\sqrt{1 + 9/49} R - 3/7 R + 1/4(3/7) R = \sqrt{58/49} R - 3/7 R + 3/28 R \approx 1.08 R - 0.428 R + 0.107 R = 0.759 R < R$. Particle does not reach origin.
For $(C)$: At $z_0 = R/\sqrt{3}$,$U_{total}(z_0) = \frac{q\sigma}{2\epsilon_0}(\sqrt{R^2 + R^2/3} - R/\sqrt{3} + 1/4(R/\sqrt{3})) = \frac{q\sigma}{2\epsilon_0}(2R/\sqrt{3} - R/\sqrt{3} + R/4\sqrt{3}) = \frac{q\sigma}{2\epsilon_0}(5R/4\sqrt{3}) > R$. It will pass the origin.
Correct option is $(A)$.

(D) The total potential difference $V_P - V_R$ is the sum of the potential differences due to the wire and the spherical shell.
$1$. Potential difference due to the infinitely long wire:
The electric field at a distance $x$ from the wire is $E = \frac{2k\lambda}{x}$,where $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$ and $\lambda = 5 \times 10^{-9} \text{ C/m}$.
The potential difference $V_P - V_R = \int_{0.5}^{2} E \, dx = \int_{0.5}^{2} \frac{2k\lambda}{x} \, dx = 2k\lambda \ln\left(\frac{2}{0.5}\right) = 2k\lambda \ln(4) = 4k\lambda \ln(2)$.
Substituting the values: $V_P - V_R = 4 \times (9 \times 10^9) \times (5 \times 10^{-9}) \times 0.7 = 180 \times 0.7 = 126 \text{ V}$.
$2$. Potential difference due to the spherical shell:
Point $P$ is at distance $0.5 \text{ m}$ (inside the shell of radius $R_s = 1 \text{ m}$),and point $R$ is at distance $2 \text{ m}$ (outside the shell).
The potential inside a shell is constant: $V_P = \frac{kQ}{R_s} = \frac{(9 \times 10^9) \times (10 \times 10^{-9})}{1} = 90 \text{ V}$.
The potential outside a shell at distance $r$ is $V_R = \frac{kQ}{r} = \frac{(9 \times 10^9) \times (10 \times 10^{-9})}{2} = 45 \text{ V}$.
Thus,$V_P - V_R = 90 - 45 = 45 \text{ V}$.
Total potential difference: $V_P - V_R = 126 + 45 = 171 \text{ V}$.

(D) The electric field $E$ for different charge distributions is as follows:
$(A)$ For a charged ring along its axis,the field is zero at the center,increases to a maximum at $r = R/\sqrt{2}$,and then decreases. This corresponds to graph $(P)$.
$(B)$ For a uniformly charged solid sphere,the field $E \propto r$ for $r < R$ and $E \propto 1/r^2$ for $r > R$. This corresponds to graph $(Q)$.
$(C)$ For a uniformly charged spherical shell,the field is zero for $r < R$ and $E \propto 1/r^2$ for $r > R$. This corresponds to graph $(R)$.
$(D)$ For a dipole (combination of $+Q$ and $-Q$),the field on the perpendicular bisector is maximum at the center and decreases as distance increases. This corresponds to graph $(S)$.
Thus,the correct matching is $(A)-(P), (B)-(Q), (C)-(R), (D)-(S)$.

(NONE) For configuration $I$: The electric field in any region between the sheets is given by $E = \frac{1}{2\epsilon_0} \sum \sigma_i$.
In region $4$,$E_4 = \frac{1}{2\epsilon_0} (\sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 + \sigma_0) = \frac{\sigma_0}{2\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0}{2\epsilon_0}$. Thus,option $C$ is incorrect.
The potential difference between the first and last sheet is $V_1 - V_6 = E_1 d + E_2 d + E_3 d + E_4 d + E_5 d$.
For configuration $I$,$E_1 = \frac{\sigma_0}{2\epsilon_0}$,$E_2 = -\frac{\sigma_0}{2\epsilon_0}$,$E_3 = \frac{\sigma_0}{2\epsilon_0}$,$E_4 = -\frac{\sigma_0}{2\epsilon_0}$,$E_5 = \frac{\sigma_0}{2\epsilon_0}$.
$V_1 - V_6 = d \frac{\sigma_0}{2\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0 d}{2\epsilon_0} = \frac{9 \times 10^{-6} \times 10^{-6}}{2 \times 9 \times 10^{-12}} = 0.5 V$. Option $B$ is incorrect.
For configuration $II$: The electric field in region $3$ is $E_3 = \frac{1}{2\epsilon_0} (\frac{\sigma_0}{2} - \sigma_0 + \sigma_0) = \frac{\sigma_0}{4\epsilon_0}$. Option $A$ is incorrect.
The potential difference between the first and last sheet for configuration $II$: $E_1 = \frac{\sigma_0}{4\epsilon_0}$,$E_2 = -\frac{\sigma_0}{4\epsilon_0}$,$E_3 = \frac{\sigma_0}{4\epsilon_0}$,$E_4 = -\frac{\sigma_0}{4\epsilon_0}$,$E_5 = \frac{\sigma_0}{4\epsilon_0}$.
$V_1 - V_6 = d (E_1 + E_2 + E_3 + E_4 + E_5) = d \frac{\sigma_0}{4\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0 d}{4\epsilon_0} \neq 0$. Option $D$ is incorrect.
Upon re-evaluating the configurations,no provided option is correct based on standard physics principles.

(B) The initial electrostatic force between two identical balls with charge $Q$ at distance $r$ is given by Coulomb's Law: $F = \frac{kQ^2}{r^2}$.
When an identical uncharged ball touches one of the charged balls,the charge $Q$ is shared equally between them. Thus,the touched ball now has charge $Q/2$ and the third ball also acquires a charge $Q/2$.
The third ball is placed at the midpoint ($r/2$ from each). Let the charges be $q_1 = Q/2$ (touched ball),$q_2 = Q$ (untouched ball),and $q_3 = Q/2$ (third ball).
The force on the third ball due to the first ball is $F_1 = \frac{k(Q/2)(Q/2)}{(r/2)^2} = \frac{kQ^2/4}{r^2/4} = \frac{kQ^2}{r^2} = F$ (directed away from the first ball).
The force on the third ball due to the second ball is $F_2 = \frac{k(Q)(Q/2)}{(r/2)^2} = \frac{kQ^2/2}{r^2/4} = \frac{2kQ^2}{r^2} = 2F$ (directed away from the second ball).
Since the forces are in opposite directions,the net force is $F_{\text{net}} = |F_2 - F_1| = |2F - F| = F$.

(C) Van de Graaff generator is an electrostatic generator which uses a moving belt to accumulate very high electric charge on a hollow metal dome.
It is designed to produce a very high potential difference (voltage),typically in the range of millions of volts.
However,the amount of charge transferred per unit time is very small,resulting in a very low current.
Therefore,it produces high voltage and low current.

(B) The Van de Graaff generator operates on two primary principles:
$1$. The phenomenon of Corona discharge at sharp points,which allows for the transfer of charge to a moving belt.
$2$. The property that charge given to a hollow conductor is transferred to its outer surface,and the potential of an isolated conductor increases as more charge is supplied to it.
Option $B$ describes the principle of a cyclotron,where electric and magnetic fields are used to accelerate charged particles. Therefore,the Van de Graaff generator is not based on the application of perpendicular electric and magnetic fields.

(A) Initial force between the spheres is given by Coulomb's law:
$F = \frac{k(4Q)(2Q)}{r^2} = \frac{8kQ^2}{r^2} \quad \dots(1)$
When the spheres are connected by a conducting wire,the total charge is redistributed equally between them:
$Q_{new} = \frac{4Q + (-2Q)}{2} = \frac{2Q}{2} = Q$
Now,the new distance is $r' = \frac{r}{2}$. The new force $F'$ is:
$F' = \frac{k(Q)(Q)}{(r/2)^2} = \frac{kQ^2}{r^2/4} = \frac{4kQ^2}{r^2}$
From equation $(1)$,we know $\frac{kQ^2}{r^2} = \frac{F}{8}$.
Substituting this into the expression for $F'$:
$F' = 4 \times \left(\frac{F}{8}\right) = \frac{F}{2}$

(C) The correct option is $C$.
When a moving electron approaches another stationary electron,the electrostatic force of repulsion between them acts in the direction opposite to the motion of the moving electron.
This repulsive force does negative work on the moving electron,which causes its kinetic energy to decrease.
According to the law of conservation of energy,the total energy of the system remains constant.
Since the kinetic energy of the moving electron decreases,its potential energy must increase as the distance between the two electrons decreases.

(B) Initial charges are $q_1 = +2 \ nC$ and $q_2 = -8 \ nC$. The magnitude of the initial attractive force is $|F| = \frac{k |q_1 q_2|}{d^2} = \frac{k (2 \times 8) \times 10^{-18}}{d^2} = \frac{16k \times 10^{-18}}{d^2}$.
When the spheres touch,the total charge is shared equally: $q_{new} = \frac{q_1 + q_2}{2} = \frac{2 - 8}{2} = -3 \ nC$. Both spheres now have a charge of $-3 \ nC$.
Let the new distance be $d'$. The new repulsive force is $|F'| = \frac{k |q_{new} q_{new}|}{(d')^2} = \frac{k (3 \times 3) \times 10^{-18}}{(d')^2} = \frac{9k \times 10^{-18}}{(d')^2}$.
Given $|F| = |F'|$,we have $\frac{16k \times 10^{-18}}{d^2} = \frac{9k \times 10^{-18}}{(d')^2}$.
$\frac{16}{d^2} = \frac{9}{(d')^2} \Rightarrow (d')^2 = \frac{9}{16} d^2$.
Taking the square root,$d' = \frac{3}{4} d$.

(A) Let $m$ be the mass of each sphere and $V$ be its volume. The density of the sphere is $\rho = m/V$,so $m = V\rho$.
In air,the forces acting on the sphere are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.
For equilibrium,$\tan(\theta/2) = \frac{F_e}{mg} = \frac{q^2}{4\pi\epsilon_0 r^2 mg}$.
In a liquid of density $\sigma$,the sphere experiences an upthrust $F_b = V\sigma g$. The effective weight is $mg' = mg - V\sigma g = Vg(\rho - \sigma)$.
The electrostatic force in the liquid is $F_e' = \frac{F_e}{K}$.
Since the angle $\theta$ remains the same,$\tan(\theta/2) = \frac{F_e'}{mg'} = \frac{F_e}{K Vg(\rho - \sigma)}$.
Equating the two expressions for $\tan(\theta/2)$:
$\frac{F_e}{V\rho g} = \frac{F_e}{K Vg(\rho - \sigma)}$.
Solving for $K$,we get $K = \frac{\rho}{\rho - \sigma}$.

(C) The initial electrostatic force between the spheres is given by Coulomb's law: $F_{\text{initial}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{|12 \times (-8)|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{96}{r^2}$.
When the spheres are brought into contact,the total charge is redistributed equally because the spheres are identical: $q_{\text{new}} = \frac{q_1 + q_2}{2} = \frac{12 + (-8)}{2} = \frac{4}{2} = 2 \mu C$.
After contact,the new electrostatic force is: $F_{\text{final}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_{\text{new}} q_{\text{new}}|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2 \times 2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4}{r^2}$.
The ratio of the magnitudes of the forces is: $\frac{|F_{\text{initial}}|}{|F_{\text{final}}|} = \frac{96/r^2}{4/r^2} = \frac{96}{4} = 24$.
Thus,the ratio is $24: 1$.

(C) Initially,the force between the two spheres is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2} = k \frac{|(+8 \mu C)(-4 \mu C)|}{r^2} = k \frac{32 \mu C^2}{r^2}$.
When the two identical conducting spheres are connected by a wire,the charges redistribute equally because they have the same radius.
The new charge on each sphere will be $q' = \frac{q_1 + q_2}{2} = \frac{+8 \mu C - 4 \mu C}{2} = +2 \mu C$.
The new force $F'$ between the spheres is $F' = k \frac{|q' q'|}{r^2} = k \frac{|(+2 \mu C)(+2 \mu C)|}{r^2} = k \frac{4 \mu C^2}{r^2}$.
Comparing the two forces: $\frac{F'}{F} = \frac{4}{32} = \frac{1}{8}$.
Therefore,$F' = \frac{F}{8}$.

(A) Let $l$ be the length of the thread. In equilibrium,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F_e$. From the geometry,$\tan \theta = \frac{F_e}{mg}$,where $\theta$ is the angle the thread makes with the vertical.
In air,the angle between the threads is $90^{\circ}$,so $\theta = 45^{\circ}$. The distance between the spheres is $r = 2l \sin 45^{\circ} = l\sqrt{2}$.
Thus,$\tan 45^{\circ} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2 mg} \Rightarrow 1 = \frac{q^2}{4\pi\epsilon_0 (2l^2) mg} \Rightarrow \frac{q^2}{4\pi\epsilon_0 l^2} = 2mg$ (Eq. $1$).
In the liquid,the angle between the threads is $60^{\circ}$,so $\theta' = 30^{\circ}$. The effective weight is $m'g = V(d - \rho_{liquid})g = V(d - \frac{2}{3}d)g = \frac{mg}{3}$. The electrostatic force is $F_e' = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q^2}{r'^2}$,where $r' = 2l \sin 30^{\circ} = l$.
Thus,$\tan 30^{\circ} = \frac{F_e'}{m'g} \Rightarrow \frac{1}{\sqrt{3}} = \frac{q^2}{4\pi\epsilon_0 \epsilon_r l^2 (mg/3)} \Rightarrow \frac{q^2}{4\pi\epsilon_0 l^2} = \frac{mg}{3\sqrt{3}} \epsilon_r$ (Eq. $2$).
Equating Eq. $1$ and Eq. $2$: $2mg = \frac{mg}{3\sqrt{3}} \epsilon_r \Rightarrow \epsilon_r = 6\sqrt{3}$.

(D) For the particles to be in equilibrium on a horizontal surface,the electrostatic repulsive force must be balanced by the maximum static frictional force.
Given: $m = 3 \ g = 3 \times 10^{-3} \ kg$,$q = 0.2 \ \mu C = 0.2 \times 10^{-6} \ C$,$r = 20 \ cm = 0.2 \ m$,$g = 10 \ ms^{-2}$,$k = 9 \times 10^9 \ Nm^2 C^{-2}$.
The electrostatic force is $F_e = \frac{k q^2}{r^2} = \frac{9 \times 10^9 \times (0.2 \times 10^{-6})^2}{(0.2)^2}$.
$F_e = \frac{9 \times 10^9 \times 0.04 \times 10^{-12}}{0.04} = 9 \times 10^{-3} \ N$.
The maximum static friction is $f_s = \mu N = \mu mg$.
Equating $F_e = f_s$,we get $\mu mg = 9 \times 10^{-3}$.
$\mu \times (3 \times 10^{-3}) \times 10 = 9 \times 10^{-3}$.
$\mu \times 3 \times 10^{-2} = 9 \times 10^{-3}$.
$\mu = \frac{9 \times 10^{-3}}{3 \times 10^{-2}} = 3 \times 10^{-1} = 0.30$.

(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 30^{\circ}$ acting downwards,the frictional force $f = \mu N$ acting upwards,and the component of the electric force $qE \cos 30^{\circ}$ acting upwards.
First,we find the normal reaction $N$ on the block:
$N = mg \cos 30^{\circ} + qE \sin 30^{\circ}$
$N = (1 \times 10 \times \frac{\sqrt{3}}{2}) + (0.01 \times 100 \times \frac{1}{2}) = 5\sqrt{3} + 0.5 \approx 8.66 + 0.5 = 9.16 \ N$
The net force $F_{net}$ along the incline is:
$F_{net} = mg \sin 30^{\circ} - \mu N - qE \cos 30^{\circ}$
$F_{net} = (1 \times 10 \times 0.5) - 0.2 \times (9.16) - (0.01 \times 100 \times \frac{\sqrt{3}}{2})$
$F_{net} = 5 - 1.832 - 0.866 = 2.302 \ N$
Since $F_{net} = ma$,and $m = 1 \ kg$,the acceleration $a = 2.302 \ ms^{-2} \approx 2.3 \ ms^{-2}$.

(B) Let the initial charges on spheres $A$ and $B$ be $q$. The distance between them is $r$.
The initial repulsive force is $F = \frac{k q^2}{r^2} = 3 \times 10^{-5} \,N$.
When uncharged sphere $C$ touches sphere $A$, the charge $q$ is shared equally between $A$ and $C$. Thus, the charge on $A$ becomes $q/2$ and the charge on $C$ becomes $q/2$.
Sphere $C$ is then placed at the midpoint between $A$ and $B$. The distance of $C$ from both $A$ and $B$ is $r/2$.
The force exerted by $A$ on $C$ is $F_{AC} = \frac{k (q/2)(q/2)}{(r/2)^2} = \frac{k q^2}{r^2} = F = 3 \times 10^{-5} \,N$ (repulsive, directed towards $B$).
The force exerted by $B$ on $C$ is $F_{BC} = \frac{k (q)(q/2)}{(r/2)^2} = 2 \frac{k q^2}{r^2} = 2F = 6 \times 10^{-5} \,N$ (repulsive, directed towards $A$).
The net force on $C$ is $F' = |F_{BC} - F_{AC}| = |2F - F| = F = 3 \times 10^{-5} \,N$.

(B) The electric field $E$ due to a charge $Q$ at distance $r$ is given by $E = \frac{k|Q|}{r^2}$. Since all charges are at the same distance $r$ from the centre,the electric field due to a charge $nq$ is proportional to $n$.
Let the electric field due to charge $-q$ at position $1$ be $\vec{E}_1$ directed towards the charge (since it is negative).
The net electric field $\vec{E}_{net}$ is the vector sum of all fields: $\vec{E}_{net} = \sum_{n=1}^{12} \vec{E}_n$.
Note that the charge at position $n$ is $-nq$. The field $\vec{E}_n$ is directed from the centre towards the position $n$.
We can pair opposite charges: $(n)$ and $(n+6)$. The net field from these two is $\vec{E}_{net, n} = \vec{E}_n + \vec{E}_{n+6} = \frac{k}{r^2} [(-n\hat{r}_n) + (-(n+6)\hat{r}_{n+6})]$.
Since $\hat{r}_{n+6} = -\hat{r}_n$,we have $\vec{E}_{net, n} = \frac{k}{r^2} [-n\hat{r}_n + (n+6)\hat{r}_n] = \frac{6k}{r^2} \hat{r}_n$.
This means the net field from each pair $(n, n+6)$ is directed towards the position $n+6$.
There are $6$ such pairs: $(1,7), (2,8), (3,9), (4,10), (5,11), (6,12)$.
The net field is $\vec{E}_{net} = \frac{6k}{r^2} [\hat{r}_7 + \hat{r}_8 + \hat{r}_9 + \hat{r}_{10} + \hat{r}_{11} + \hat{r}_{12}]$.
These unit vectors point towards $7, 8, 9, 10, 11, 12$ on the clock. The resultant of these vectors points towards the direction between $9$ and $10$,which corresponds to $9:30$.

(D) Given: $Q_1 = +2 \times 10^{-6} \ C$, $Q_2 = -4 \times 10^{-6} \ C$, distance $d = 3 \ m$. Let $P$ be at a distance $r_1$ from $Q_1$ and $r_2$ from $Q_2$. Since $P$ is between them, $r_1 + r_2 = 3 \ m$. Let $r_2 = x$, then $r_1 = 3 - x$.
Electric potential $V = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q_1}{r_1} + \frac{Q_2}{r_2} \right) = 0$.
$\frac{2 \times 10^{-6}}{3 - x} + \frac{-4 \times 10^{-6}}{x} = 0 \Rightarrow \frac{2}{3 - x} = \frac{4}{x} \Rightarrow x = 6 - 2x \Rightarrow 3x = 6 \Rightarrow x = 2 \ m$.
So, $r_2 = 2 \ m$ and $r_1 = 1 \ m$.
The electric field $E$ at $P$ due to $Q_1$ is $E_1 = \frac{k |Q_1|}{r_1^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{1^2} = 18000 \ N/C$ (directed towards $Q_2$).
The electric field $E$ at $P$ due to $Q_2$ is $E_2 = \frac{k |Q_2|}{r_2^2} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2} = 9000 \ N/C$ (directed towards $Q_2$).
Since both fields point in the same direction (towards $Q_2$), the net electric field $E = E_1 + E_2 = 18000 + 9000 = 27000 \ N/C$.

(D) Electric potential is a scalar quantity. The potential at the centre due to $9$ charges is the sum of potentials due to each charge:
$V = 9 \times \frac{Kq}{R} = \frac{9Kq}{R} \quad \dots(i)$
Electric field is a vector quantity. In a regular polygon of $10$ sides,if charges were present at all $10$ corners,the net electric field at the centre would be zero due to symmetry (each pair of opposite charges cancels out).
Let the missing charge be at the $10^{th}$ corner. If we place a charge $+q$ and $-q$ at the $10^{th}$ corner,the net field is the field due to $-q$ at that corner,as the other $10$ charges cancel out.
Thus,the magnitude of the electric field at the centre is:
$E = \frac{Kq}{R^2} \quad \dots(ii)$
Dividing Eq. $(i)$ by Eq. $(ii)$:
$\frac{V}{E} = \frac{\frac{9Kq}{R}}{\frac{Kq}{R^2}} = 9R$

(A) Given: Mass $m = 1 \ g = 10^{-3} \ kg$,distance $r = 1 \ m$,charge $q = 1 \ fC = 10^{-15} \ C$.
$1$. Gravitational force: $F_g = \frac{G m_1 m_2}{r^2} = \frac{(6.67 \times 10^{-11}) \times (10^{-3}) \times (10^{-3})}{1^2} = 6.67 \times 10^{-17} \ N$.
$2$. Electrostatic force: $F_e = \frac{k q_1 q_2}{r^2} = \frac{(9 \times 10^9) \times (10^{-15}) \times (10^{-15})}{1^2} = 9 \times 10^{-21} \ N$.
Comparing the two,$F_g > F_e$ $(6.67 \times 10^{-17} \ N > 9 \times 10^{-21} \ N)$.
Therefore,the gravitational force is the dominant force.

(A) The electric field $E_{x}$ due to two positive point charges $q_{1}$ and $q_{2}$ on the $x$-axis is given by the superposition principle: $E_{x} = \frac{k q_{1}}{x^{2}} + \frac{k q_{2}}{(x-10)^{2}}$ for $x < 0$,$E_{x} = \frac{k q_{1}}{x^{2}} - \frac{k q_{2}}{(x-10)^{2}}$ for $0 < x < 10$,and $E_{x} = -\frac{k q_{1}}{x^{2}} - \frac{k q_{2}}{(x-10)^{2}}$ for $x > 10$.
$1$. For $x < 0$: Both charges contribute to a negative electric field in the $x$-direction,so $E_{x} < 0$. As $x \to 0^{-}$,$E_{x} \to -\infty$.
$2$. For $0 < x < 10$: The field due to $q_{1}$ is positive and the field due to $q_{2}$ is negative. There is a point between the charges where $E_{x} = 0$. As $x \to 0^{+}$,$E_{x} \to +\infty$. As $x \to 10^{-}$,$E_{x} \to -\infty$.
$3$. For $x > 10$: Both charges contribute to a positive electric field in the $x$-direction,so $E_{x} > 0$. As $x \to 10^{+}$,$E_{x} \to +\infty$. As $x \to \infty$,$E_{x} \to 0$.
Comparing these characteristics with the given options,the graph in option $A$ correctly represents these behaviors.

(C) According to the principle of conservation of energy,the initial total energy of the system must equal the final total energy.
Since the particles are released from rest,the initial kinetic energy $KE_i = 0$.
The initial potential energy $PE_i$ is the sum of the electrostatic potential energies of all pairs of charges.
In a square of side $a$,there are $4$ pairs of charges at distance $a$ (sides) and $2$ pairs of charges at distance $\sqrt{2}a$ (diagonals).
$PE_i = 4 \times \frac{kq^2}{a} + 2 \times \frac{kq^2}{\sqrt{2}a} = \frac{kq^2}{a} (4 + \sqrt{2})$.
When the particles are infinitely far apart,the final potential energy $PE_f = 0$.
Thus,$KE_f = PE_i = \frac{kq^2}{a} (4 + \sqrt{2})$.

(C) Analysis of statements:
$A$. Incorrect: Electrostatic field lines originate from positive charges and terminate at negative charges; they do not form closed loops (unlike magnetic field lines).
$B$. Correct: For a positive charge $(q > 0)$,the electric field lines point radially outward.
$C$. Correct: Gauss's Law is a direct consequence of the inverse-square nature of Coulomb's law.
$D$. Correct: The electrostatic force is a conservative force,so the work done in moving a charge along a closed path is zero.
$E$. Correct: Coulomb's force is a central force,and motion under a central force is always confined to a plane.
Therefore,statements $B, C, D,$ and $E$ are correct.

(B) The force of gravity on the oil droplet is balanced by the electric force acting upwards. The droplet must be attracted towards the top plate $A$. The force of gravity is $F_g = mg = (\rho \cdot \frac{4}{3}\pi r^3)g$.
The electric force is $F_e = qE = q(\frac{V_A - V_B}{d})$.
Setting $F_g = F_e$,we have $\rho \cdot \frac{4}{3}\pi r^3 g = q \frac{\Delta V}{d}$.
Given $\rho = 1500 \text{ kg/m}^3$,$r = 10^{-3} \text{ m}$,$g = 10 \text{ m/s}^2$,$d = \frac{0.12}{\pi} \text{ m}$,$q = 5 \times 10^{-9} \text{ C}$.
Substituting values: $1500 \times \frac{4}{3} \times \pi \times (10^{-3})^3 \times 10 = 5 \times 10^{-9} \times \frac{\Delta V}{(0.12/\pi)}$.
$20 \times 10^{-6} \times \pi = 5 \times 10^{-9} \times \frac{\Delta V \times \pi}{0.12}$.
Solving for $\Delta V$: $\Delta V = \frac{20 \times 10^{-6} \times 0.12}{5 \times 10^{-9}} = 480 \text{ V}$.
Since $V_A - V_B = 480 \text{ V}$,only option $(B)$ $580 \text{ V} - 100 \text{ V} = 480 \text{ V}$ satisfies this condition.