Electrostatic Force and Coulombs Law Questions in English

(D) The force between two point electric charges is given by Coulomb's law.
According to Coulomb's law,the magnitude of the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The mathematical expression is $F = k \frac{|q_1 q_2|}{r^2}$,where $k$ is Coulomb's constant.
Therefore,the correct option is $D$.

(D) According to Coulomb's Law,the electrostatic force $F$ between two point charges is inversely proportional to the square of the distance $r$ between them: $F \propto \frac{1}{r^2}$.
If the initial distance is $r$,the initial force is $F = k \frac{q_1 q_2}{r^2}$.
When the distance is halved,the new distance becomes $r' = \frac{r}{2}$.
The new force $F'$ is given by $F' = k \frac{q_1 q_2}{(r/2)^2} = k \frac{q_1 q_2}{r^2 / 4} = 4 \left( k \frac{q_1 q_2}{r^2} \right) = 4F$.
Therefore,the force becomes four times the original force.

(B) According to Coulomb's Law,the force exerted by charge $q_1$ on $q_2$ is equal in magnitude and opposite in direction to the force exerted by $q_2$ on $q_1$ (Newton's Third Law of Motion).
Therefore,the magnitude of the force on both charges is the same.
The ratio of the forces acting on them is $1:1$.

(C) According to the principle of superposition,the force exerted by one charge on another is independent of the presence of other charges in the vicinity.
The force between $q_1$ and $q_2$ is given by Coulomb's Law: $F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
This force depends only on the magnitudes of $q_1$ and $q_2$ and the distance $r$ between them. The presence of a third charge $q_3$ introduces a new force on $q_2$ (due to $q_3$),but it does not alter the force exerted by $q_1$ on $q_2$.
Therefore,the force of $q_1$ exerted on $q_2$ remains unchanged.

(B) According to Coulomb's Law,the force between two charges $q_1$ and $q_2$ separated by a distance $r$ in air is given by $F_a = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.
In a medium with dielectric constant $K$,the force is given by $F_m = \frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{r^2}$.
Therefore,the ratio of the forces is $\frac{F_a}{F_m} = \frac{\frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}}{\frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{r^2}} = K$.
Thus,the ratio $F_a : F_m = K : 1$.

(C) Let a test charge $q_0$ be placed at the centre $O$.
$1$. The charges at $A$ $(+q)$ and $C$ $(+q)$ are equal and equidistant from $O$. The force due to $A$ is repulsive and directed towards $C$,while the force due to $C$ is repulsive and directed towards $A$. Since these forces are equal in magnitude and opposite in direction,they cancel each other out.
$2$. The charges at $B$ $(+2q)$ and $D$ $(-2q)$ are equidistant from $O$. The force due to $B$ $(+2q)$ on $q_0$ is repulsive (directed away from $B$ towards $D$). The force due to $D$ $(-2q)$ on $q_0$ is attractive (directed towards $D$).
$3$. Since both forces are directed towards $D$,the resultant force is non-zero and directed along the diagonal $BD$ towards $D$.

(A) Let the positions of the charges be $x_1 = 0$,$x_2 = l/2$,and $x_3 = l$.
The force on charge $q$ at $x = l$ due to charge $4q$ at $x = 0$ is given by Coulomb's Law:
$F_1 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(4q)(q)}{l^2} = \frac{4q^2}{4\pi\varepsilon_0 l^2}$
The force on charge $q$ at $x = l$ due to charge $Q$ at $x = l/2$ is:
$F_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Q)(q)}{(l/2)^2} = \frac{4Qq}{4\pi\varepsilon_0 l^2}$
For the resultant force on $q$ to be zero,the sum of forces must be zero:
$F_1 + F_2 = 0$
$\frac{4q^2}{4\pi\varepsilon_0 l^2} + \frac{4Qq}{4\pi\varepsilon_0 l^2} = 0$
$4q^2 + 4Qq = 0$
$4Qq = -4q^2$
$Q = -q$

(A) In the absence of gravity,the only force acting on the spheres is the electrostatic repulsive force between them.
Since both spheres have the same charge $+Q$,they repel each other.
To maximize the distance between them and reach an equilibrium position,the threads will stretch out in opposite directions,forming a straight line.
Thus,the angle between the two threads is $180^\circ$.
The distance between the two charges is $r = L + L = 2L$.
According to Coulomb's law,the electrostatic force $F$ between them is $F = \frac{1}{4\pi \varepsilon_0} \frac{Q \cdot Q}{(2L)^2} = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{(2L)^2}$.
This force is equal to the tension $T$ in each thread.

(A) According to Coulomb's Law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:
$F = \frac{k q_1 q_2}{r^2}$
Given:
$q_1 = q_2 = 1\;C$
$r = 1\;km = 1000\;m = 10^3\;m$
$k = 9 \times 10^9\;N\cdot m^2/C^2$
Substituting the values:
$F = \frac{(9 \times 10^9) \times (1) \times (1)}{(10^3)^2}$
$F = \frac{9 \times 10^9}{10^6}$
$F = 9 \times 10^3\;N$

(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = \frac{k Q_1 Q_2}{r^2}$.
Initially,$F = 12\,N$,so $12 = \frac{k(2)(6)}{r^2} = \frac{12k}{r^2}$,which implies $\frac{k}{r^2} = 1$.
After adding $-2\,C$ to each charge,the new charges are $Q_1' = 2 - 2 = 0\,C$ and $Q_2' = 6 - 2 = 4\,C$.
The new force $F'$ is $F' = \frac{k Q_1' Q_2'}{r^2} = \frac{k(0)(4)}{r^2} = 0\,N$.

(B) Let the side length of the square be $a$. The charges are equal,so let $q_1 = q_2 = q_3 = q$.
$F_{12}$ is the force between two adjacent corners,so the distance between them is $a$. Thus,$F_{12} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{a^2}$.
$F_{13}$ is the force between two opposite corners (diagonal),so the distance between them is $a\sqrt{2}$. Thus,$F_{13} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{(a\sqrt{2})^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{2a^2}$.
Taking the ratio: $\frac{F_{12}}{F_{13}} = \frac{\frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{a^2}}{\frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{2a^2}} = \frac{1/a^2}{1/2a^2} = 2$.

(C) In the $CGS$ system,Coulomb's law is given by $F = \frac{q_1 q_2}{r^2}$.
The force on charge $B$ $(+12\,e.s.u.)$ due to charge $A$ $(+15\,e.s.u.)$ is repulsive and acts along $BA$ (downwards in the diagram). Its magnitude is:
$F_A = \frac{15 \times 12}{3^2} = \frac{180}{9} = 20\,dyne$.
The force on charge $B$ $(+12\,e.s.u.)$ due to charge $C$ $(-20\,e.s.u.)$ is attractive and acts along $BC$ (towards $C$). Its magnitude is:
$F_C = \frac{|12 \times (-20)|}{4^2} = \frac{240}{16} = 15\,dyne$.
Since the angle between $BA$ and $BC$ is $90^\circ$,the net force on $B$ is:
$F_{net} = \sqrt{F_A^2 + F_C^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\,dyne$.

(D) The electrostatic force between two charges in a vacuum is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When a medium with dielectric constant $K$ is introduced,the force becomes $F' = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{d^2}$.
Therefore,the new force is $F' = \frac{F}{K}$.
Given $K = 4$,the new force is $F' = \frac{F}{4}$.

(A) For point charges,the force is given by Coulomb's law: $F_e = \frac{k Q^2}{d^2}$.
When the charges are placed on spheres of radius $R = 0.3 \, d$,the distance between the centers is $d$.
Since the spheres are close to each other ($R$ is a significant fraction of $d$),the electrostatic induction effect causes the charges to redistribute on the surfaces of the spheres.
Specifically,the positive charge on one sphere is attracted towards the negative charge on the other sphere,causing the charges to shift closer to each other on the inner faces of the spheres.
Because the effective distance between the centers of charge becomes less than $d$,the force of attraction increases.
Therefore,the new force is greater than $F_e$.

(A) The electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ in a medium with dielectric constant $K$ is given by $F_m = \frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{r^2}$.
In air,the dielectric constant $K_{air} \approx 1$,so the force is $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.
When the ions are placed in water,the new force $F'$ is given by $F' = \frac{1}{4\pi\varepsilon_0 K} \frac{q_1 q_2}{r^2}$.
Comparing the two expressions,we get $F' = \frac{F}{K}$.

(C) According to Coulomb's Law,the force between two charges is given by $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
Rearranging for the permittivity of free space $\varepsilon_0$,we get $\varepsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$.
The value of the constant $\frac{1}{4 \pi \varepsilon_0}$ is approximately $9 \times 10^9 \, N \cdot m^2 / C^2$.
Therefore,$\varepsilon_0 = \frac{1}{4 \pi \times 9 \times 10^9} \approx 8.854 \times 10^{-12} \, C^2 / (N \cdot m^2)$.

(C) Initially, the force between the two spheres with charges $+q$ and $-q$ separated by distance $r$ is given by Coulomb's Law: $F = k\frac{q^2}{r^2}$.
When a third sphere with charge $+q$ is placed at the midpoint (distance $r/2$ from both), it experiences two forces:
$1$. Force due to the $+q$ charge at $A$: $F_A = k\frac{q \cdot q}{(r/2)^2} = 4k\frac{q^2}{r^2} = 4F$ (directed away from $A$, towards $C$).
$2$. Force due to the $-q$ charge at $C$: $F_C = k\frac{q \cdot q}{(r/2)^2} = 4k\frac{q^2}{r^2} = 4F$ (directed towards $C$).
The net force on the middle sphere is $F_{net} = F_A + F_C = 4F + 4F = 8F$.
Since both forces are directed towards the $-q$ charge, the net force is $8F$ towards the $-q$ charge.

(A) The Coulomb force between two charges $q$ and $Q-q$ separated by a distance $r$ is given by:
$F = \frac{k q (Q - q)}{r^2}$
To find the condition for maximum force,we differentiate $F$ with respect to $q$ and set it to zero:
$\frac{dF}{dq} = \frac{k}{r^2} \frac{d}{dq} (Qq - q^2) = 0$
$Q - 2q = 0$
$Q = 2q$
Therefore,the ratio $\frac{Q}{q} = 2$.

(A) The force between two charges $q_1$ and $q_2$ separated by a distance $r$ in air is given by $F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
When the medium is replaced by a dielectric of constant $k$,the permittivity of the medium becomes $\varepsilon = k \varepsilon_0$.
The new force $F'$ is given by $F' = \frac{1}{4\pi \varepsilon} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi k \varepsilon_0} \frac{q_1 q_2}{r^2}$.
Therefore,$F' = \frac{F}{k} = k^{-1} F$.
Since $k > 1$ for any dielectric medium,the force decreases and becomes $k^{-1}$ times the original force.

(B) According to Coulomb's Law,the electrostatic force $F$ between two point charges is inversely proportional to the square of the distance $r$ between them: $F \propto \frac{1}{r^2}$.
Given initial distance $r_1 = 0.06\,m$ and initial force $F_1 = 5\,N$.
When each charge is moved towards the other by $0.01\,m$,the new distance $r_2 = 0.06\,m - 0.01\,m - 0.01\,m = 0.04\,m$.
Using the ratio formula: $\frac{F_1}{F_2} = \left( \frac{r_2}{r_1} \right)^2$.
Substituting the values: $\frac{5}{F_2} = \left( \frac{0.04}{0.06} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.
Therefore,$F_2 = 5 \times \frac{9}{4} = \frac{45}{4} = 11.25\,N$.

(A) According to Coulomb's Law,the force between two charges in a vacuum or air is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When the charges are immersed in a medium with dielectric constant $K$,the force becomes $F_m = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{d^2}$.
Therefore,the relationship between the force in the medium and the force in air is $F_m = \frac{F}{K}$.
Given that $K = 2$,the new force is $F_m = \frac{F}{2}$.

(A) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{Q_1 Q_2}{r^2}$.
Initially,$Q_1 = +3\,\mu C$ and $Q_2 = +8\,\mu C$,with force $F = 40\,N$.
After adding $-5\,\mu C$ to each charge,the new charges are:
$Q_1' = 3\,\mu C - 5\,\mu C = -2\,\mu C$
$Q_2' = 8\,\mu C - 5\,\mu C = +3\,\mu C$
Since the distance $r$ remains the same,the ratio of the forces is $\frac{F'}{F} = \frac{Q_1' Q_2'}{Q_1 Q_2}$.
Substituting the values: $\frac{F'}{40} = \frac{(-2) \times (3)}{3 \times 8} = \frac{-6}{24} = -\frac{1}{4}$.
Thus,$F' = 40 \times (-1/4) = -10\,N$.
The negative sign indicates that the force is now attractive.

(B) The side length $r = 10\,cm = 0.1\,m$. The charges are $q_A = 1 \times 10^{-6}\,C$, $q_B = -1 \times 10^{-6}\,C$, and $q_C = 2 \times 10^{-6}\,C$.
The force $F_A$ on $C$ due to charge at $A$ is repulsive:
$F_A = \frac{k |q_A q_C|}{r^2} = \frac{9 \times 10^9 \times 10^{-6} \times 2 \times 10^{-6}}{(0.1)^2} = 1.8\,N$.
The force $F_B$ on $C$ due to charge at $B$ is attractive:
$F_B = \frac{k |q_B q_C|}{r^2} = \frac{9 \times 10^9 \times 10^{-6} \times 2 \times 10^{-6}}{(0.1)^2} = 1.8\,N$.
The angle between the force vectors $F_A$ and $F_B$ is $120^\circ$ because the internal angle of the equilateral triangle is $60^\circ$ and the vectors are oriented such that the angle between them is $180^\circ - 60^\circ = 120^\circ$.
The resultant force $F_{net}$ is given by:
$F_{net} = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos(120^\circ)}$
$F_{net} = \sqrt{(1.8)^2 + (1.8)^2 + 2(1.8)(1.8)(-0.5)}$
$F_{net} = \sqrt{1.8^2 + 1.8^2 - 1.8^2} = 1.8\,N$.

(B) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{Q_1 Q_2}{R^2}$.
Initially,the charges are $Q_1 = +10\,\mu C$ and $Q_2 = -20\,\mu C$. The force is $F_1 = k \frac{(10)(-20)}{R^2} = -200 \frac{k}{R^2}$.
When the spheres are brought into contact,the total charge is redistributed equally because the spheres have equal radii. The new charge on each sphere is $Q' = \frac{Q_1 + Q_2}{2} = \frac{10 - 20}{2} = -5\,\mu C$.
After separation to the same distance $R$,the new force is $F_2 = k \frac{(-5)(-5)}{R^2} = 25 \frac{k}{R^2}$.
The ratio of the forces is $\frac{F_1}{F_2} = \frac{-200 \frac{k}{R^2}}{25 \frac{k}{R^2}} = \frac{-200}{25} = -8$.
Thus,the ratio $F_1 : F_2$ is $-8:1$.

(A) According to Coulomb's Law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given by:
$F = k \cdot \frac{q_1 q_2}{r^2}$
Where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
Given: $q_1 = q_2 = 2 \times 10^{-6} \, C$ and $r = 0.5 \, m$.
Substituting the values:
$F = 9 \times 10^9 \times \frac{(2 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.5)^2}$
$F = 9 \times 10^9 \times \frac{4 \times 10^{-12}}{0.25}$
$F = 9 \times 10^9 \times 16 \times 10^{-12}$
$F = 144 \times 10^{-3} = 0.144 \, N$.

(C) The force between two point charges $q_1$ and $q_2$ separated by a distance $d$ in a vacuum is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When a conducting plate (like copper) of thickness $t$ is placed between the charges,the effective distance between the charges is reduced. The effective distance $d_{eff}$ is given by $d_{eff} = (d - t) + t\sqrt{K}$,where $K$ is the dielectric constant.
For a conductor,the dielectric constant $K = \infty$.
Therefore,$d_{eff} = (d - t) + t\sqrt{\infty} = \infty$.
Since the force $F \propto \frac{1}{d_{eff}^2}$,as $d_{eff} \to \infty$,the force $F \to 0$.

(A) The Coulomb force between two charges is given by $F = k \frac{|q_1 q_2|}{r^2}$.
Here,$k = 9 \times 10^9 \, N \cdot m^2/C^2$,$q_1 = q_2 = e = 1.6 \times 10^{-19} \, C$,and $r = 1 \, \mathring{A} = 10^{-10} \, m$.
Substituting the values:
$F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(10^{-10})^2}$
$F = 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{10^{-20}}$
$F = 9 \times 2.56 \times 10^{9 - 38 + 20}$
$F = 23.04 \times 10^{-9} = 2.304 \times 10^{-8} \, N$.
Thus,the correct option is $A$.

(C) $1$. Calculate the number of atoms in $10\,g$ of copper: $N = \frac{\text{mass}}{\text{atomic weight}} \times N_A = \frac{10}{63.5} \times 6.022 \times 10^{23} \approx 9.48 \times 10^{22}$ atoms.
$2$. Calculate the number of electrons transferred: $n = \frac{N}{10^6} = \frac{9.48 \times 10^{22}}{10^6} = 9.48 \times 10^{16}$ electrons.
$3$. Calculate the charge on each ball: $Q = n \times e = 9.48 \times 10^{16} \times 1.6 \times 10^{-19} \approx 0.015\,C$.
$4$. Calculate the Coulomb force using $F = k \frac{Q_1 Q_2}{r^2}$,where $k = 9 \times 10^9\,N\cdot m^2/C^2$,$Q_1 = Q_2 = 0.015\,C$,and $r = 0.1\,m$: $F = 9 \times 10^9 \times \frac{(0.015)^2}{(0.1)^2} = 9 \times 10^9 \times \frac{2.25 \times 10^{-4}}{0.01} = 9 \times 10^9 \times 2.25 \times 10^{-2} = 20.25 \times 10^7 \approx 2.0 \times 10^8\,N$.

(A) Let the three charges $q$ be placed at vertices $A, B,$ and $C$ of an equilateral triangle of side $L$. Let a charge $Q$ be placed at the centroid $O$ of the triangle.
The distance from each vertex to the centroid $O$ is $r = \frac{L}{\sqrt{3}}$.
The electrostatic force exerted by each charge $q$ on the charge $Q$ at the center is given by Coulomb's Law: $F = \frac{1}{4\pi \varepsilon _0} \frac{qQ}{r^2} = \frac{1}{4\pi \varepsilon _0} \frac{qQ}{(L/\sqrt{3})^2} = \frac{3}{4\pi \varepsilon _0} \frac{qQ}{L^2}$.
Since the charges are identical and placed symmetrically at the vertices of an equilateral triangle,the three force vectors $\overrightarrow{F_A}, \overrightarrow{F_B},$ and $\overrightarrow{F_C}$ acting on $Q$ have equal magnitudes and are directed away from (or towards) the vertices at angles of $120^\circ$ to each other.
The vector sum of three equal forces acting at $120^\circ$ to each other is zero. Therefore,the net electrostatic force on the charge $Q$ at the center is zero.

(C) Let the charges at $A, B,$ and $C$ be $+Q, -Q,$ and $+Q$ respectively. The distance between any two vertices is $a$.
The force on charge at $A$ due to charge at $B$ is $F_B = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (attractive,directed towards $B$).
The force on charge at $A$ due to charge at $C$ is $F_C = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (repulsive,directed away from $C$).
Resolving these forces into components parallel and normal to $BC$:
$1$. The components of $F_B$ and $F_C$ parallel to $BC$ are $F_B \cos 60^\circ$ and $F_C \cos 60^\circ$ respectively,both directed towards the left.
$2$. The components of $F_B$ and $F_C$ normal to $BC$ are $F_B \sin 60^\circ$ (downwards) and $F_C \sin 60^\circ$ (upwards) respectively.
Since $|F_B| = |F_C|$,the net force in the direction normal to $BC$ is $F_C \sin 60^\circ - F_B \sin 60^\circ = 0$.

(D) For the particles to experience no net force,the gravitational force of attraction must be balanced by the electrostatic force of repulsion.
$|\vec{F_G}| = |\vec{F_e}|$
$G \frac{m^2}{r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q^2}{r^2}$
Canceling $r^2$ from both sides,we get:
$G m^2 = \frac{1}{4\pi \varepsilon_0} q^2$
Rearranging the terms to find $\frac{q}{m}$:
$\frac{q^2}{m^2} = 4\pi \varepsilon_0 G$
$\frac{q}{m} = \sqrt{4\pi \varepsilon_0 G}$

(C) The Coulomb force between the nucleus (charge $+e$) and the electron (charge $-e$) is given by Coulomb's Law: $\overrightarrow{F} = K \frac{q_1 q_2}{r^2} \hat{r}$.
Substituting $q_1 = +e$ and $q_2 = -e$,we get $\overrightarrow{F} = K \frac{(+e)(-e)}{r^2} \hat{r} = -K \frac{e^2}{r^2} \hat{r}$.
Since the unit vector $\hat{r} = \frac{\vec{r}}{r}$,we can substitute this into the expression:
$\overrightarrow{F} = -K \frac{e^2}{r^2} \left( \frac{\vec{r}}{r} \right) = -K \frac{e^2}{r^3} \vec{r}$.

(C) The force between two point charges $Q_1$ and $Q_2$ in air at a distance $r$ is given by $F = \frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2}$.
When the same charges are placed in a medium of dielectric constant $k$ at a distance $r'$,the force $F'$ is given by $F' = \frac{1}{4\pi\varepsilon_0 k} \frac{Q_1 Q_2}{r'^2}$.
Given that the force remains the same,$F = F'$,so $\frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2} = \frac{1}{4\pi\varepsilon_0 k} \frac{Q_1 Q_2}{r'^2}$.
Simplifying the equation,we get $\frac{1}{r^2} = \frac{1}{k r'^2}$.
Rearranging for $r'$,we get $r'^2 = \frac{r^2}{k}$,which implies $r' = \frac{r}{\sqrt{k}}$.

(C) Let $k = \frac{1}{4\pi \varepsilon_0}$. The charge at $B$ experiences forces from charges at $A, C,$ and $D$.
$1$. Force due to charge at $A$: $F_A = \frac{kq^2}{a^2}$ (directed along $AB$ extended).
$2$. Force due to charge at $C$: $F_C = \frac{kq^2}{a^2}$ (directed along $CB$ extended).
$3$. Force due to charge at $D$: $F_D = \frac{kq^2}{(a\sqrt{2})^2} = \frac{kq^2}{2a^2}$ (directed along $DB$ extended).
Since $F_A$ and $F_C$ are perpendicular,their resultant $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{\left(\frac{kq^2}{a^2}\right)^2 + \left(\frac{kq^2}{a^2}\right)^2} = \sqrt{2} \frac{kq^2}{a^2}$.
This resultant $F_{AC}$ acts in the same direction as $F_D$ (along the diagonal $DB$).
Therefore,the net force $F_{net} = F_{AC} + F_D = \sqrt{2} \frac{kq^2}{a^2} + \frac{kq^2}{2a^2} = \frac{kq^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{kq^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.
Substituting $k = \frac{1}{4\pi \varepsilon_0}$,we get $F_{net} = \left( \frac{1 + 2\sqrt{2}}{2} \right) \frac{q^2}{4\pi \varepsilon_0 a^2}$.

(D) Initially,the force between $B$ and $C$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2}$.
When the uncharged conductor (let's call it $D$) is brought in contact with $B$,the charge $Q$ is shared equally between them because they have equal radii. Thus,the charge on $B$ becomes $Q_B = Q/2$.
Next,the conductor $D$ (now carrying charge $Q/2$) is brought in contact with $C$ (which has charge $Q$). The total charge $(Q/2 + Q) = 3Q/2$ is shared equally between $C$ and $D$. Thus,the new charge on $C$ becomes $Q_C = (3Q/2) / 2 = 3Q/4$.
The new force of repulsion between $B$ and $C$ is $F' = k \frac{Q_B \cdot Q_C}{r^2} = k \frac{(Q/2) \cdot (3Q/4)}{r^2} = \frac{3}{8} \left( k \frac{Q^2}{r^2} \right) = \frac{3}{8} F$.

(A) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{q_1 q_2}{r^2}$.
Initially,the charges are $q_1 = +7\,\mu C$ and $q_2 = -5\,\mu C$. The magnitude of the force is $F = k \frac{(7)(5)}{r^2} = k \frac{35}{r^2}$.
When an additional charge of $-2\,\mu C$ is added to each sphere,the new charges become:
$q_1' = +7\,\mu C - 2\,\mu C = +5\,\mu C$
$q_2' = -5\,\mu C - 2\,\mu C = -7\,\mu C$
The new force $F'$ is given by $F' = k \frac{|q_1' q_2'|}{r^2} = k \frac{|(5)(-7)|}{r^2} = k \frac{35}{r^2}$.
Comparing the two expressions,we find that $F' = F$.

(A) The gravitational force between an electron and a proton is given by $F_G = \frac{G m_e m_p}{r^2}$.
Substituting the values: $F_G = \frac{6.7 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{(5 \times 10^{-11})^2} \approx 3.9 \times 10^{-47} \, N$.
The electrostatic force between an electron and a proton is given by $F_e = \frac{1}{4\pi \epsilon_0} \frac{e^2}{r^2}$.
Substituting the values: $F_e = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(5 \times 10^{-11})^2} \approx 9.22 \times 10^{-8} \, N$.
The ratio of electrostatic to gravitational force is $\frac{F_e}{F_G} = \frac{9.22 \times 10^{-8}}{3.9 \times 10^{-47}} \approx 2.36 \times 10^{39}$.

(D) According to Coulomb's Law,the force between two point charges is $F = k \frac{Q_1 Q_2}{r^2}$. Since the distance $r$ remains constant,$F \propto Q_1 Q_2$.
Initial charges: $Q_1 = 3 \times 10^{-6} \, C$,$Q_2 = 8 \times 10^{-6} \, C$.
Initial force: $F_1 = 6 \times 10^{-3} \, N$.
New charges after adding $-6 \times 10^{-6} \, C$ to each:
$Q_1' = 3 \times 10^{-6} - 6 \times 10^{-6} = -3 \times 10^{-6} \, C$
$Q_2' = 8 \times 10^{-6} - 6 \times 10^{-6} = 2 \times 10^{-6} \, C$
Using the ratio: $\frac{F_2}{F_1} = \frac{Q_1' Q_2'}{Q_1 Q_2}$
$\frac{F_2}{6 \times 10^{-3}} = \frac{(-3 \times 10^{-6}) \times (2 \times 10^{-6})}{(3 \times 10^{-6}) \times (8 \times 10^{-6})}$
$\frac{F_2}{6 \times 10^{-3}} = \frac{-6}{24} = -\frac{1}{4}$
$F_2 = -\frac{6 \times 10^{-3}}{4} = -1.5 \times 10^{-3} \, N$.
The negative sign indicates that the force is attractive.

(A) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = k\frac{Q^2}{r^2}$.
When sphere $C$ (uncharged) is brought in contact with sphere $A$ (charge $Q$),the charge redistributes equally between them. Thus,the new charge on $A$ is $Q_A = Q/2$ and on $C$ is $Q_C = Q/2$. Sphere $B$ remains with charge $Q_B = Q$.
Sphere $C$ is placed at the midpoint,so the distance from $A$ is $r/2$ and from $B$ is $r/2$.
The force on $C$ due to $A$ is $F_A = k\frac{(Q/2)(Q/2)}{(r/2)^2} = k\frac{Q^2/4}{r^2/4} = k\frac{Q^2}{r^2} = F$. This force acts away from $A$ (repulsive).
The force on $C$ due to $B$ is $F_B = k\frac{(Q)(Q/2)}{(r/2)^2} = k\frac{Q^2/2}{r^2/4} = 2k\frac{Q^2}{r^2} = 2F$. This force acts away from $B$ (repulsive).
Since $F_A$ and $F_B$ act in opposite directions,the net force on $C$ is $F_{net} = |F_B - F_A| = |2F - F| = F$.

(D) According to Coulomb's Law,the force between two charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{Q_1 Q_2}{r^2}$.
Given that the charges are equal in magnitude,let $Q_1 = Q_2 = Q$. Thus,the initial force is $F = k \frac{Q^2}{r^2}$.
When the charges are halved,the new charges become $Q' = Q/2$. When the distance is doubled,the new distance becomes $r' = 2r$.
The new force $F'$ is given by $F' = k \frac{(Q/2)(Q/2)}{(2r)^2}$.
$F' = k \frac{Q^2 / 4}{4r^2} = \frac{1}{16} \left( k \frac{Q^2}{r^2} \right)$.
Since $F = k \frac{Q^2}{r^2}$,we have $F' = F / 16$.

(B) The total force $F$ acting on the $1 \, C$ charge at the origin due to the infinite series of charges is given by Coulomb's Law:
$F = \sum_{i=0}^{\infty} \frac{k q_1 q_2}{r_i^2} = \frac{1}{4\pi \varepsilon_0} \left( \frac{1 \times 10^{-6} \times 1}{1^2} + \frac{1 \times 10^{-6} \times 1}{2^2} + \frac{1 \times 10^{-6} \times 1}{4^2} + \frac{1 \times 10^{-6} \times 1}{8^2} + ... \infty \right)$
$F = (9 \times 10^9) \times 10^{-6} \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ... \infty \right)$
The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/4$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
$F = 9 \times 10^3 \times \frac{4}{3} = 3 \times 10^3 \times 4 = 12000 \, N$.

(B) Coulomb's law describes the electrostatic force between charged particles.
$A$,$C$,and $D$ involve electrostatic interactions (attraction between electrons and nuclei,or between atoms/molecules).
However,the force that binds protons and neutrons (nucleons) within the nucleus is the strong nuclear force,which is not an electric force.
Therefore,the statement in option $B$ is incorrect.

(B) For the system of three charges to be in equilibrium,the net force on each charge must be zero.
Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $x$. The charge $q$ is placed at the midpoint $C$ (distance $x/2$ from both $A$ and $B$).
Consider the equilibrium of charge $Q$ at point $B$. The force exerted by charge $Q$ at $A$ on charge $Q$ at $B$ must be balanced by the force exerted by charge $q$ at $C$ on charge $Q$ at $B$.
$F_{AB} + F_{CB} = 0$
$\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(x/2)^2} = 0$
$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0$
$Q^2 + 4qQ = 0$
$4qQ = -Q^2$
$q = -\frac{Q}{4}$

(D) At the neutral point,the electric field due to both charges must be equal in magnitude and opposite in direction.
Let the distance of the neutral point from the $20 \, C$ charge be $r_1 = 20 \, cm = 0.2 \, m$.
The total distance is $60 \, cm = 0.6 \, m$.
Therefore,the distance of the neutral point from charge $Q$ is $r_2 = 60 \, cm - 20 \, cm = 40 \, cm = 0.4 \, m$.
Equating the electric field magnitudes:
$E_1 = E_2$
$\frac{k \cdot 20}{r_1^2} = \frac{k \cdot Q}{r_2^2}$
$\frac{20}{(0.2)^2} = \frac{Q}{(0.4)^2}$
$\frac{20}{0.04} = \frac{Q}{0.16}$
$Q = 20 \times \frac{0.16}{0.04}$
$Q = 20 \times 4 = 80 \, C$.

(B) In the balance condition,the electric force equals the gravitational force:
$QE = mg$
Since $E = \frac{V}{d}$ and $m = \frac{4}{3}\pi r^3 \rho$,we have:
$Q \frac{V}{d} = \frac{4}{3}\pi r^3 \rho g$
This implies $Q \propto \frac{r^3}{V}$.
For the two drops,we can write:
$\frac{Q_1}{Q_2} = \left( \frac{r_1}{r_2} \right)^3 \times \frac{V_2}{V_1}$
Given $Q_1 = Q$,$r_1 = r$,$V_1 = 2400\,V$,$r_2 = \frac{r}{2}$,and $V_2 = 600\,V$:
$\frac{Q}{Q_2} = \left( \frac{r}{r/2} \right)^3 \times \frac{600}{2400}$
$\frac{Q}{Q_2} = (2)^3 \times \frac{1}{4} = 8 \times \frac{1}{4} = 2$
Therefore,$Q_2 = \frac{Q}{2}$.

(C) First, we calculate the net force on each charge:
$1$. For the charge at $x=0$ $(+4q)$: The force due to $-q$ at $x=a$ is attractive, and the force due to $+4q$ at $x=2a$ is repulsive. The net force is $F = k \frac{(4q)(q)}{a^2} - k \frac{(4q)(4q)}{(2a)^2} = \frac{4kq^2}{a^2} - \frac{4kq^2}{a^2} = 0$.
$2$. For the charge at $x=a$ $(-q)$: The force due to $+4q$ at $x=0$ is attractive, and the force due to $+4q$ at $x=2a$ is also attractive. The net force is $F = k \frac{(4q)(q)}{a^2} - k \frac{(4q)(q)}{a^2} = 0$.
$3$. For the charge at $x=2a$ $(+4q)$: By symmetry, the net force is $0$.
Since the net force on all charges is zero, they are in equilibrium. However, if any charge is displaced slightly, the restoring force does not act to return it to its original position; instead, the net force increases, pushing it further away. Thus, all charges are in unstable equilibrium.