Electrostatic Force and Coulombs Law Questions in English

(B) In equilibrium,the forces acting on one of the spherical balls are the tension $T$ in the thread,the electrostatic force $F_e$,and the gravitational force $mg$.
Let the length of the thread be $L = 1\,m$ and the angle with the vertical be $\theta = 30^o$.
The distance $r$ between the two charges is $r = 2L \sin(30^o) = 2 \times 1 \times 0.5 = 1\,m$.
The electrostatic force is $F_e = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{r^2} = 9 \times 10^9 \times \frac{(10 \times 10^{-6})^2}{1^2} = 9 \times 10^9 \times 10^{-10} = 0.9\,N$.
For equilibrium in the horizontal direction: $T \sin(30^o) = F_e$.
$T \times 0.5 = 0.9$.
$T = 1.8\,N$.

(B) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$. For the system to be in equilibrium,the net force on charge $q$ must be zero.
Let $F_1$ be the force due to charge $+e$ and $F_2$ be the force due to charge $+9e$.
For equilibrium,$|F_1| = |F_2|$.
Using Coulomb's law: $\frac{kqe}{(16-x)^2} = \frac{kq(9e)}{x^2}$.
Taking the square root on both sides: $\frac{1}{16-x} = \frac{3}{x}$.
Solving for $x$: $x = 3(16-x) \implies x = 48 - 3x \implies 4x = 48 \implies x = 12\, cm$.
Thus,the charge $q$ should be placed at $12\, cm$ from the $+9e$ charge.

(C) For a third charge $q$ to experience no net force,the forces exerted by the two charges $Q_1$ and $Q_2$ must be equal in magnitude and opposite in direction. Since the charges have opposite signs,the equilibrium point must lie outside the line segment joining them,closer to the charge with the smaller magnitude $(-2.7 \times 10^{-11} \, C)$.
Let the third charge $q$ be placed at a distance $x$ from the charge $Q_2 = -2.7 \times 10^{-11} \, C$. The distance from $Q_1 = 5 \times 10^{-11} \, C$ is $(x + 0.2) \, m$.
Equating the magnitudes of the forces: $\frac{k |Q_1| |q|}{(x + 0.2)^2} = \frac{k |Q_2| |q|}{x^2}$.
Substituting the values: $\frac{5 \times 10^{-11}}{(x + 0.2)^2} = \frac{2.7 \times 10^{-11}}{x^2}$.
$\frac{5}{(x + 0.2)^2} = \frac{2.7}{x^2} \implies \frac{\sqrt{5}}{x + 0.2} = \frac{\sqrt{2.7}}{x}$.
$2.236 x = 1.643 (x + 0.2) \implies 2.236 x = 1.643 x + 0.3286$.
$0.593 x = 0.3286 \implies x \approx 0.554 \, m$. Given the options,the closest value is $0.556 \, m$.

(D) The force on charge $Q$ at $x = l$ is due to the charges $4Q$ at $x = 0$ and $q$ at $x = l/2$.
Using Coulomb's Law,the force exerted by $4Q$ on $Q$ is $F_1 = \frac{1}{4\pi\varepsilon_0} \frac{(4Q)(Q)}{l^2}$.
The force exerted by $q$ on $Q$ is $F_2 = \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(l/2)^2}$.
For the net force on $Q$ to be zero,$F_1 + F_2 = 0$.
$\frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{l^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{l^2/4} = 0$.
$\frac{4Q^2}{l^2} + \frac{4qQ}{l^2} = 0$.
Dividing by $\frac{4Q}{l^2}$ (assuming $Q \neq 0$ and $l \neq 0$),we get $Q + q = 0$.
Therefore,$q = -Q$.

(C) Let the two equal charges be $Q$ and the third charge be $q$. The distance between the two charges $Q$ is $d$. The third charge $q$ is placed at a distance $x$ on the perpendicular bisector.
The force exerted by each charge $Q$ on $q$ is $F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qq}{x^2 + (d/2)^2}$.
The net force $F_{net}$ on $q$ is the sum of the components of these forces along the perpendicular bisector:
$F_{net} = 2F \cos \theta$,where $\cos \theta = \frac{x}{\sqrt{x^2 + (d/2)^2}}$.
Substituting the values:
$F_{net} = 2 \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qq}{x^2 + d^2/4} \cdot \frac{x}{(x^2 + d^2/4)^{1/2}} = \frac{2Qqx}{4\pi\varepsilon_0 (x^2 + d^2/4)^{3/2}}$.
For $F_{net}$ to be maximum,$\frac{dF_{net}}{dx} = 0$.
Using the quotient rule or differentiating $f(x) = x(x^2 + a^2)^{-3/2}$ where $a = d/2$:
$f'(x) = (x^2 + a^2)^{-3/2} + x \cdot (-3/2)(x^2 + a^2)^{-5/2} \cdot 2x = 0$.
$(x^2 + a^2)^{-5/2} [ (x^2 + a^2) - 3x^2 ] = 0$.
$a^2 - 2x^2 = 0 \implies x^2 = a^2/2 \implies x = a/\sqrt{2}$.
Since $a = d/2$,we get $x = \frac{d/2}{\sqrt{2}} = \frac{d}{2\sqrt{2}}$.

(C) Let $F_2$ be the force applied by $+q_2$ on $-q_1$. Since they have opposite signs,the force is attractive and acts along the positive $x$-axis. Thus,$F_2 = k \frac{q_1 q_2}{b^2}$.
Let $F_3$ be the force applied by $-q_3$ on $-q_1$. Since they have the same sign,the force is repulsive. The force $F_3$ acts along the line joining them,making an angle $\theta$ with the negative $y$-axis. The magnitude is $F_3 = k \frac{q_1 q_3}{a^2}$.
The $x$-component of $F_3$ is $F_{3x} = F_3 \sin \theta = k \frac{q_1 q_3}{a^2} \sin \theta$,acting in the positive $x$-direction.
The net $x$-component of the force on $-q_1$ is $F_x = F_2 + F_{3x} = k \frac{q_1 q_2}{b^2} + k \frac{q_1 q_3}{a^2} \sin \theta$.
Therefore,$F_x = k q_1 \left( \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin \theta \right)$.
Thus,$F_x \propto \left( \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin \theta \right)$.

(A) According to the method of electrical images,an earthed conducting plane acts like an image charge of equal magnitude but opposite sign placed at an equal distance behind the plane.
Here,the charge $q = 40 \ statC$ is at a distance $d = 2 \ cm$ from the plate.
The image charge $q' = -40 \ statC$ is at a distance $d = 2 \ cm$ behind the plate.
The total distance between the charge and its image is $r = 2d = 4 \ cm$.
Using Coulomb's law in the $CGS$ system,the force $F$ is given by:
$F = \frac{|q \cdot q'|}{r^2} = \frac{40 \times 40}{4^2} = \frac{1600}{16} = 100 \ dynes$.

(B) The electrical force of attraction between the electron and proton is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.
Here,$k = 9 \times 10^9 \ N \cdot m^2/C^2$,$q_1 = q_2 = 1.6 \times 10^{-19} \ C$,and $r = 2.5 \times 10^{-11} \ m$.
Substituting the values:
$F = \frac{9 \times 10^9 \times (1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(2.5 \times 10^{-11})^2}$
$F = \frac{9 \times 2.56 \times 10^{9 - 19 - 19}}{6.25 \times 10^{-22}}$
$F = \frac{23.04 \times 10^{-29}}{6.25 \times 10^{-22}}$
$F = 3.6864 \times 10^{-7} \ N \approx 3.7 \times 10^{-7} \ N$.

(A) In a $CsCl$ crystal,the $Cs^{+}$ ions are located at the eight corners of a cube of side length $a$.
The $Cl^{-}$ ion is placed at the exact centre of this cube.
Due to the symmetry of the $bcc$ structure,each $Cs^{+}$ ion at a corner exerts an electrostatic force on the central $Cl^{-}$ ion.
For every $Cs^{+}$ ion,there is an identical $Cs^{+}$ ion located diametrically opposite to it with respect to the centre.
The forces exerted by these pairs of $Cs^{+}$ ions on the central $Cl^{-}$ ion are equal in magnitude and opposite in direction.
According to the principle of superposition,the vector sum of all these forces is zero.
Therefore,the net electrostatic force on the $Cl^{-}$ ion is zero.

(B) The charge of the nucleus of a hydrogen atom is $q_1 = +e$.
The charge of the electron is $q_2 = -e$.
The Coulomb force $\vec{F}$ is given by the formula $\vec{F} = \frac{k q_1 q_2}{r^2} \hat{r}$,where $\hat{r}$ is the unit vector pointing from the nucleus to the electron.
Substituting the values,we get $\vec{F} = \frac{k(e)(-e)}{r^2} \hat{r} = -\frac{k e^2}{r^2} \hat{r}$.
Since $\hat{r} = \frac{\vec{r}}{r}$,we can rewrite the expression as $\vec{F} = -\frac{k e^2}{r^2} \left( \frac{\vec{r}}{r} \right) = -k \frac{e^2}{r^3} \vec{r}$.

(C) Let the distance between two spheres $A$ $(+q)$ and $B$ $(-q)$ be $r$. The initial force between them is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}$.
When a third sphere $O$ with charge $+q$ is placed at their midpoint (at distance $r/2$):
Repulsive force exerted by sphere $A$ $(+q)$ on sphere $O$ $(+q)$: $F_{AO} = \frac{1}{4 \pi \varepsilon_0} \frac{q \cdot q}{(r/2)^2} = 4 \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \right) = 4F$ (towards $-q$).
Attractive force exerted by sphere $B$ $(-q)$ on sphere $O$ $(+q)$: $F_{OB} = \frac{1}{4 \pi \varepsilon_0} \frac{|-q| \cdot q}{(r/2)^2} = 4 \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \right) = 4F$ (towards $-q$).
Total force $F' = F_{AO} + F_{OB} = 4F + 4F = 8F$. This force will act in the direction of the $-q$ charge.

(A) According to Coulomb's Law,the force between two charges is given by $F = \frac{k q_1 q_2}{r^2}$.
Initially,$q_1 = 2 \, \mu \text{C}$ and $q_2 = 6 \, \mu \text{C}$,and $F = 12 \, \text{N}$.
So,$12 = \frac{k (2 \times 10^{-6}) (6 \times 10^{-6})}{r^2} \implies \frac{k}{r^2} = \frac{12}{12 \times 10^{-12}} = 10^{12} \, \text{N} \cdot \text{m}^2/\text{C}^2$.
After adding $-4 \, \mu \text{C}$ to each charge:
$q_1' = 2 - 4 = -2 \, \mu \text{C}$
$q_2' = 6 - 4 = +2 \, \mu \text{C}$
The new force $F'$ is $F' = \frac{k q_1' q_2'}{r^2} = \frac{k (-2 \times 10^{-6}) (2 \times 10^{-6})}{r^2}$.
Substituting $\frac{k}{r^2} = 10^{12}$,we get $F' = 10^{12} \times (-4 \times 10^{-12}) = -4 \, \text{N}$.
The negative sign indicates that the force is attractive. Thus,the magnitude is $4 \, \text{N}$ and the nature is attractive.

(D) The charges at $B$ and $C$ are both $q$ and the distance between them and $A$ is $a$. The force exerted by charge $B$ on $A$ is $F_{AB} = \frac{kq^2}{a^2}$ along the line $BA$ extended. The force exerted by charge $C$ on $A$ is $F_{AC} = \frac{kq^2}{a^2}$ along the line $CA$ extended. Since the triangle is equilateral,the angle of these forces with the vertical (perpendicular to $BC$) is $30^\circ$. The horizontal components of these forces cancel each other out. The vertical components add up: $F_{net} = F_{AB} \cos(30^\circ) + F_{AC} \cos(30^\circ) = 2 \times \frac{kq^2}{a^2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}kq^2}{a^2}$.

(D) Let the third charge $q$ be placed at a distance $x$ from the charge $9e$. Then,its distance from the charge $16e$ is $(70 - x) \ cm$.
For the third charge to be in equilibrium,the net electrostatic force on it must be zero. Therefore,the magnitude of the force exerted by $9e$ must equal the magnitude of the force exerted by $16e$:
$F_1 = F_2$
$\frac{k \cdot (9e) \cdot q}{x^2} = \frac{k \cdot (16e) \cdot q}{(70 - x)^2}$
$\frac{9}{x^2} = \frac{16}{(70 - x)^2}$
Taking the square root on both sides:
$\frac{3}{x} = \frac{4}{70 - x}$
$3(70 - x) = 4x$
$210 - 3x = 4x$
$7x = 210$
$x = 30 \ cm$
Thus,the charge should be placed at a distance of $30 \ cm$ from $9e$. The distance from $16e$ is $70 - 30 = 40 \ cm$. Therefore,both options $(a)$ and $(b)$ are correct.

(B) Step $1$: Calculate the distances between the charges. Let the side length of the square be $a$. The diagonal length is $AC = \sqrt{a^2 + a^2} = \sqrt{2}a$. The distance from the center $O$ to any corner is $OC = \frac{AC}{2} = \frac{a}{\sqrt{2}}$.
Step $2$: Equilibrium condition. For the system to be in equilibrium,the net force on every charge must be zero. By symmetry,the net force on the central charge $q$ is zero regardless of its value. Thus,we must ensure the net force on any one of the corner charges (e.g.,at point $C$) is zero.
Step $3$: Forces acting on the charge at corner $C$. Let the charges at corners $A, B, C, D$ be $-Q$. The force due to $B$ is $F_1 = \frac{KQ^2}{a^2}$ (repulsive),the force due to $D$ is $F_2 = \frac{KQ^2}{a^2}$ (repulsive),and the force due to $A$ is $F_4 = \frac{KQ^2}{(\sqrt{2}a)^2} = \frac{KQ^2}{2a^2}$ (repulsive). The force due to central charge $q$ is $F_3 = \frac{K|q|Q}{(a/\sqrt{2})^2} = \frac{2K|q|Q}{a^2}$ (attractive,so $q$ must be positive).
Step $4$: Summing the forces along the diagonal $OC$. The resultant of $F_1$ and $F_2$ is $\sqrt{F_1^2 + F_2^2} = \sqrt{2} \frac{KQ^2}{a^2}$. For equilibrium: $F_3 + F_4 + \sqrt{2}F_1 = 0$. Since $F_3$ is attractive (towards center) and others are repulsive (away from center),we set: $\frac{2KqQ}{a^2} + \frac{KQ^2}{2a^2} + \sqrt{2}\frac{KQ^2}{a^2} = 0$.
Step $5$: Solving for $q$: $2q + \frac{Q}{2} + \sqrt{2}Q = 0 \Rightarrow 2q = -Q(\frac{1}{2} + \sqrt{2}) \Rightarrow q = -\frac{Q}{4}(1 + 2\sqrt{2})$. However,checking the force direction,for the corner charge to be in equilibrium,$q$ must be positive to attract the corner charge. Re-evaluating the sign convention,the correct magnitude is $q = \frac{Q}{4}(1 + 2\sqrt{2})$.

(A) According to the principle of superposition in electrostatics,the force exerted by one point charge on another is independent of the presence of other charges in the vicinity.
The force $F$ between two point charges $q_1$ and $q_2$ is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$,where $r$ is the distance between them.
Since the force $F$ depends only on the magnitudes of $q_1$ and $q_2$ and the distance $r$ between them,the introduction of a third charge $q_3$ does not alter the individual force exerted by $q_1$ on $q_2$.
Although the net force on $q_2$ will change due to the additional force exerted by $q_3$,the specific force exerted by $q_1$ on $q_2$ remains $F$.

(A) According to Coulomb's Law,the electrostatic force between two point charges $Q_1$ and $Q_2$ separated by a distance $r$ is given by $F = k \frac{|Q_1 Q_2|}{r^2}$.
Newton's third law of motion states that for every action,there is an equal and opposite reaction.
Therefore,the force exerted by charge $Q_1$ on $Q_2$ is equal in magnitude and opposite in direction to the force exerted by charge $Q_2$ on $Q_1$.
Since the magnitudes of the forces are equal,the ratio of the forces exerted by them on each other is $1 : 1$.

(D) The force between two charges is a central force,which is conservative in nature.
Work done by a conservative force in a closed path is zero.
Alternatively,the work done $W$ is given by the integral $W = \int \vec{F} \cdot d\vec{r}$.
In this case,the force $\vec{F}$ is directed radially outward (along the line joining the charges),and the displacement $d\vec{r}$ is along the tangent to the circle.
Since the force is always perpendicular to the displacement (angle $\theta = 90^\circ$),the dot product $\vec{F} \cdot d\vec{r} = F \cdot dr \cdot \cos(90^\circ) = 0$.
Therefore,the total work done is $0$.

(C) The electric permittivity of free space, denoted by $\varepsilon_0$, is a fundamental physical constant.
From 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 $\varepsilon_0$, we get $\varepsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$.
The $SI$ unit of $\varepsilon_0$ is $C^2/(N \cdot m^2)$ or $F/m$.
The experimentally determined value is approximately $8.854 \times 10^{-12} \, C^2/(N \cdot m^2)$.

(A) In a satellite,the condition is one of weightlessness,meaning the effective gravitational force $mg = 0$.
Due to the electrostatic force of repulsion between the two identical negative charges $Q$,the balls will push each other away until the strings are fully extended in opposite directions.
Since the strings are of length $L$,the distance between the two balls becomes $2L$.
The angle between the two strings is $180^{\circ}$.
The tension $T$ in each string is balanced by the electrostatic force of repulsion $F_e$ between the two charges.
Using Coulomb's Law: $F_e = \frac{k |Q_1 Q_2|}{r^2} = \frac{k Q^2}{(2L)^2} = \frac{k Q^2}{4L^2}$.
Therefore,the tension $T = F_e = \frac{k Q^2}{4L^2} \text{ N}$.

(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:
$F = k \frac{|q_1 q_2|}{r^2}$
Given that the initial distance is $r$ and the new distance is $r' = 2r$.
The new force $F'$ is given by:
$F' = k \frac{|q_1 q_2|}{(r')^2} = k \frac{|q_1 q_2|}{(2r)^2}$
$F' = k \frac{|q_1 q_2|}{4r^2} = \frac{1}{4} \left( k \frac{|q_1 q_2|}{r^2} \right)$
$F' = \frac{F}{4}$
Therefore,the electrostatic force becomes one-fourth of its original value.

(B) Step $1$: Equilibrium of the central charge $Q$.
For the system to be in equilibrium,the net force on each charge must be zero.
By symmetry,the two outer charges $q$ exert equal and opposite forces on $Q$,which cancel each other out. Thus,the net force on $Q$ is zero regardless of the value of $q$.
Step $2$: Equilibrium of the outer charge $q$.
Let the distance between the two charges $q$ be $2x$. The distance from $q$ to $Q$ is $x$.
The net force on the charge $q$ at the right must be zero:
$F = \frac{kQq}{x^2} + \frac{kq^2}{(2x)^2} = 0$
Dividing by $k/x^2$ (assuming $x \neq 0$):
$Qq + \frac{q^2}{4} = 0$
$Qq = -\frac{q^2}{4}$
$Q = -\frac{q}{4}$
Therefore,$q = -4Q$.

(A) Step $1$: Analyze the symmetry of the $CsCl$ crystal structure.
In the $bcc$ structure of $CsCl$,the $Cl^-$ ion is located at the exact center of the cube,while the eight $Cs^+$ ions are located at the eight corners of the cube.
Step $2$: Calculate the net force.
Due to the cubic symmetry,each $Cs^+$ ion at a corner exerts an electrostatic force on the central $Cl^-$ ion. Since the $Cl^-$ ion is at the geometric center,for every $Cs^+$ ion at one corner,there is an identical $Cs^+$ ion at the diagonally opposite corner.
Step $3$: Conclusion.
The forces exerted by these pairs of diagonally opposite $Cs^+$ ions on the central $Cl^-$ ion are equal in magnitude and opposite in direction. Consequently,these forces cancel each other out completely.
Therefore,the net electrostatic force on the $Cl^-$ ion is $zero$.

(C) Let the side of the square be $a$. Place the charges $Q$ at $(0, a)$ and $(a, 0)$,and charges $q$ at $(0, 0)$ and $(a, a)$.
Consider the force on charge $Q$ at $(a, 0)$.
The forces acting on $Q$ are:
$1$. Force due to $q$ at $(0, 0)$: $\vec{F}_1 = \frac{kQq}{a^2} \hat{i}$
$2$. Force due to $q$ at $(a, a)$: $\vec{F}_2 = \frac{kQq}{a^2} \hat{j}$
$3$. Force due to $Q$ at $(0, a)$: $\vec{F}_3 = \frac{kQ^2}{(\sqrt{2}a)^2} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right)$
For the net force to be zero,the sum of components must be zero:
$F_x = \frac{kQq}{a^2} + \frac{kQ^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = 0$
$\frac{kQq}{a^2} = -\frac{kQ^2}{2\sqrt{2}a^2}$
$q = -\frac{Q}{2\sqrt{2}}$
Therefore,$\frac{Q}{q} = -2\sqrt{2}$.

(D) For the drop to remain stationary,the upward electric force must balance the downward gravitational force.
$qE = mg$
Given:
$E = 100 \ V m^{-1}$
$m = 1.6 \times 10^{-3} \ g = 1.6 \times 10^{-6} \ kg$
$g \approx 10 \ m s^{-2}$
$q = \frac{mg}{E} = \frac{1.6 \times 10^{-6} \times 10}{100} = 1.6 \times 10^{-7} \ C$
The number of electrons $n$ is given by $n = \frac{q}{e}$,where $e = 1.6 \times 10^{-19} \ C$.
$n = \frac{1.6 \times 10^{-7}}{1.6 \times 10^{-19}} = 10^{12}$.

(A) Given: Force $F = 0.144 \, N$, distance $r = 5 \, cm = 5 \times 10^{-2} \, m$, and charges are identical $(q_1 = q_2 = q)$.
According to Coulomb's Law: $F = \frac{k q^2}{r^2}$.
Substituting the values: $0.144 = \frac{(9 \times 10^9) \times q^2}{(5 \times 10^{-2})^2}$.
$0.144 = \frac{9 \times 10^9 \times q^2}{25 \times 10^{-4}}$.
$q^2 = \frac{0.144 \times 25 \times 10^{-4}}{9 \times 10^9}$.
$q^2 = 0.016 \times 25 \times 10^{-13} = 0.4 \times 10^{-13} = 4 \times 10^{-14}$.
$q = \sqrt{4 \times 10^{-14}} = 2 \times 10^{-7} \, C = 0.2 \times 10^{-6} \, C = 0.2 \, \mu C$.

(D) Let $\rho$ be the density of the sphere and $\sigma$ be the density of the liquid. When the spheres are in air,the equilibrium condition is $\tan \theta = \frac{F_e}{mg}$.
When immersed in a liquid,the effective weight becomes $mg' = V(\rho - \sigma)g$ and the electrostatic force becomes $F_e' = \frac{F_e}{K}$,where $K$ is the dielectric constant.
The equilibrium condition in the liquid is $\tan \theta = \frac{F_e'}{m'g} = \frac{F_e / K}{V(\rho - \sigma)g}$.
Since the angle $\theta$ remains the same,we equate the two expressions for $\tan \theta$:
$\frac{F_e}{V\rho g} = \frac{F_e}{K V(\rho - \sigma)g}$.
This simplifies to $K = \frac{\rho}{\rho - \sigma}$.
Given $\rho = 1.6 \, g \cdot cm^{-3}$ and $\sigma = 0.8 \, g \cdot cm^{-3}$,we get $K = \frac{1.6}{1.6 - 0.8} = \frac{1.6}{0.8} = 2$.

(B) According to Coulomb's Law,the force in air is given by $F_{air} = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.
The force in a medium (oil) is given by $F_{medium} = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{d^2}$,where $\epsilon_r$ is the relative permittivity.
Given that the force remains the same $(F_{air} = F_{medium})$,we have:
$\frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{d^2}$.
Simplifying the equation,we get $r^2 = \epsilon_r d^2$,which implies $d = \frac{r}{\sqrt{\epsilon_r}}$.
Given $r = 50 \ cm$ and $\epsilon_r = 5$,we calculate:
$d = \frac{50}{\sqrt{5}} = \frac{50}{2.236} \approx 22.36 \ cm$.
Thus,the distance in oil is approximately $22.3 \ cm$.

(C) The charge $Q$ is at the midpoint between $-q$ and $+q$. The distance from each charge to $Q$ is $r/2$.
The force exerted by $-q$ on $Q$ is $F_A = \frac{kQq}{(r/2)^2} = \frac{4kQq}{r^2}$ (attractive force towards $-q$).
The force exerted by $+q$ on $Q$ is $F_B = \frac{kQq}{(r/2)^2} = \frac{4kQq}{r^2}$ (repulsive force away from $+q$).
Since both forces act in the same direction (towards the left,towards $-q$),the net force is:
$F_{net} = F_A + F_B = \frac{4kQq}{r^2} + \frac{4kQq}{r^2} = \frac{8kQq}{r^2}$.

(A) Initially,the force between the two spheres with charges $q$ and $q$ separated by distance $r$ is given by Coulomb's Law:
$F = \frac{kq^2}{r^2}$
When $50\%$ of the charge from one sphere is transferred to the other,the new charges become:
$q_1 = q - 0.5q = 0.5q = \frac{q}{2}$
$q_2 = q + 0.5q = 1.5q = \frac{3q}{2}$
The new force $F'$ is:
$F' = \frac{k q_1 q_2}{r^2} = \frac{k (q/2) (3q/2)}{r^2}$
$F' = \frac{3}{4} \frac{kq^2}{r^2}$
Since $F = \frac{kq^2}{r^2}$,we get:
$F' = \frac{3}{4} F = 0.75 F$

(A) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$. For the system to be in equilibrium,the net force on the charge $q$ must be zero.
Let $F_1$ be the force due to charge $+e$ and $F_2$ be the force due to charge $+9e$.
For equilibrium,$|F_1| = |F_2|$.
$\frac{kqe}{(16-x)^2} = \frac{kq(9e)}{x^2}$
$\frac{1}{(16-x)^2} = \frac{9}{x^2}$
Taking the square root on both sides:
$\frac{1}{16-x} = \frac{3}{x}$
$x = 3(16-x)$
$x = 48 - 3x$
$4x = 48$
$x = 12\, cm$ from $+9e$.

(C) The two spheres are each charged with $+q$ and are separated by a distance $r = 2a$.
According to Coulomb's Law,the electrostatic repulsive force $F$ between the two charges is given by:
$F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}$
Substituting the given values $q_1 = q$,$q_2 = q$,and $r = 2a$:
$F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q \cdot q}{(2a)^2}$
$F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q^2}{4a^2}$
$F = \frac{q^2}{16\pi \varepsilon_0 a^2}$
Since the spheres are in equilibrium,the tension $T$ in the string must balance the electrostatic repulsive force.
Therefore,$T = F = \frac{q^2}{16\pi \varepsilon_0 a^2}$.

(B) Let the charge $4q$ be at position $x=0$ and charge $q$ be at position $x=l$. The charge $Q$ is placed at $x=l/2$.
For the net force on charge $q$ to be zero,the force exerted by $4q$ on $q$ must be equal and opposite to the force exerted by $Q$ on $q$.
The force exerted by $4q$ on $q$ is $F_1 = \frac{k(4q)(q)}{l^2}$.
The force exerted by $Q$ on $q$ is $F_2 = \frac{k(Q)(q)}{(l/2)^2}$.
For the net force to be zero,$F_1 + F_2 = 0$,which implies $F_1 = -F_2$.
$\frac{k(4q)(q)}{l^2} = -\frac{k(Q)(q)}{(l/2)^2}$.
$\frac{4q}{l^2} = -\frac{Q}{l^2/4}$.
$4q = -4Q$.
Therefore,$Q = -q$.

(B) Let the charges be $q_1 = q_2 = q_3 = q$.
Let the side of the square be $a$.
The distance between adjacent corners (e.g.,$1$ and $2$) is $r_{12} = a$.
The distance between opposite corners (e.g.,$1$ and $3$) is $r_{13} = \sqrt{a^2 + a^2} = a\sqrt{2}$.
Using Coulomb's Law,the force $F_{12} = \frac{k q_1 q_2}{r_{12}^2} = \frac{k q^2}{a^2}$.
The force $F_{13} = \frac{k q_1 q_3}{r_{13}^2} = \frac{k q^2}{(a\sqrt{2})^2} = \frac{k q^2}{2a^2}$.
Therefore,the ratio $\frac{F_{12}}{F_{13}} = \frac{k q^2 / a^2}{k q^2 / 2a^2} = 2$.

(A) Given: Mass $m = 1 \, g = 10^{-3} \, kg$,Charge $q = 10^{-9} \, C$,Distance $x = 0.3 \, cm = 3 \times 10^{-3} \, m$,$g \approx 10 \, m/s^2$.
Electrostatic force $F_e = \frac{k q^2}{x^2} = \frac{9 \times 10^9 \times (10^{-9})^2}{(3 \times 10^{-3})^2} = \frac{9 \times 10^{-9}}{9 \times 10^{-6}} = 10^{-3} \, N$.
Gravitational force $mg = 10^{-3} \times 10 = 10^{-2} \, N$.
For equilibrium,$\tan \theta = \frac{F_e}{mg} = \frac{10^{-3}}{10^{-2}} = 0.1$.
Therefore,$\theta = \tan^{-1}(0.1)$.

(A) Each carbon atom has $6$ electrons and $6$ protons. Total number of electrons (and protons) = $6 \times 10^{23}$.
The total charge $q$ at each pole is $q = (6 \times 10^{23}) \times (1.6 \times 10^{-19} \ C) = 9.6 \times 10^4 \ C$.
The distance between the North Pole and the South Pole is the diameter of the Earth,$d = 2R = 2 \times 6400 \ km = 1.28 \times 10^7 \ m$.
Using Coulomb's Law: $F = \frac{k q_1 q_2}{d^2} = \frac{(9 \times 10^9) \times (9.6 \times 10^4)^2}{(1.28 \times 10^7)^2}$.
$F = \frac{9 \times 10^9 \times 92.16 \times 10^8}{1.6384 \times 10^{14}} = \frac{829.44 \times 10^{17}}{1.6384 \times 10^{14}} \approx 506 \times 10^3 \ N = 5.06 \times 10^5 \ N$.
Thus,the force is approximately $5 \times 10^5 \ N$.

(A) Let the radii of the spheres be $R_1 = 3 \ cm = 3 \times 10^{-2} \ m$ and $R_2 = 1 \ cm = 1 \times 10^{-2} \ m$. The potential $V$ of each sphere is $10 \ V$.
The charge $Q$ on a sphere of radius $R$ at potential $V$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$,so $Q = 4\pi\epsilon_0 R V$.
For the first sphere: $Q_1 = \frac{R_1 V}{k} = \frac{3 \times 10^{-2} \times 10}{9 \times 10^9} = \frac{1}{3} \times 10^{-10} \ C$.
For the second sphere: $Q_2 = \frac{R_2 V}{k} = \frac{1 \times 10^{-2} \times 10}{9 \times 10^9} = \frac{1}{9} \times 10^{-10} \ C$.
The distance between the centers is $d = 10 \ cm = 0.1 \ m$.
The electrostatic force $F$ is given by Coulomb's Law: $F = k \frac{Q_1 Q_2}{d^2}$.
$F = (9 \times 10^9) \times \frac{(\frac{1}{3} \times 10^{-10}) \times (\frac{1}{9} \times 10^{-10})}{(0.1)^2} = (9 \times 10^9) \times \frac{\frac{1}{27} \times 10^{-20}}{10^{-2}} = \frac{9}{27} \times 10^9 \times 10^{-18} = \frac{1}{3} \times 10^{-9} \ N$.

(B) The position vectors are $\vec{r}_A = (\hat{i} + 3\hat{j} + 2\hat{k}) \ m$ and $\vec{r}_B = (3\hat{i} + 5\hat{j} + \hat{k}) \ m$.
The displacement vector from $B$ to $A$ is $\vec{r} = \vec{r}_A - \vec{r}_B = (1-3)\hat{i} + (3-5)\hat{j} + (2-1)\hat{k} = -2\hat{i} - 2\hat{j} + \hat{k} \ m$.
The distance $r$ between the charges is $r = |\vec{r}| = \sqrt{(-2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \ m$.
Using Coulomb's law,the magnitude of the force is $F = \frac{k |q_1 q_2|}{r^2}$.
Substituting the values: $F = \frac{9 \times 10^9 \times (5 \times 10^{-6}) \times (3 \times 10^{-6})}{3^2}$.
$F = \frac{9 \times 10^9 \times 15 \times 10^{-12}}{9} = 15 \times 10^{-3} \ N = 1.5 \times 10^{-2} \ N$.

(D) Given $Q = Q_1 + Q_2$,so $Q_2 = Q - Q_1$.
The electrostatic force between them is given by Coulomb's Law: $F = \frac{k Q_1 Q_2}{R^2} = \frac{k Q_1 (Q - Q_1)}{R^2} = \frac{k}{R^2} (Q Q_1 - Q_1^2)$.
For the force to be maximum,the derivative of $F$ with respect to $Q_1$ must be zero: $\frac{dF}{dQ_1} = 0$.
$\frac{d}{dQ_1} [\frac{k}{R^2} (Q Q_1 - Q_1^2)] = \frac{k}{R^2} (Q - 2Q_1) = 0$.
Since $\frac{k}{R^2} \neq 0$,we have $Q - 2Q_1 = 0$,which implies $Q_1 = \frac{Q}{2}$.
Substituting this back,$Q_2 = Q - Q_1 = Q - \frac{Q}{2} = \frac{Q}{2}$.
Thus,the force is maximum when $Q_1 = Q_2 = \frac{Q}{2}$.

(A) At equilibrium,the forces acting on each ball are tension $T$,electrostatic force $F_e$,and weight $mg$.
Resolving forces: $F_e = T \sin \theta$ $(i)$ and $mg = T \cos \theta$ (ii).
Dividing $(i)$ by (ii),we get $\tan \theta = \frac{F_e}{mg} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$.
Given $\tan \theta \approx \sin \theta = \frac{x/2}{L} = \frac{x}{2L}$.
Equating the two expressions for $\tan \theta$: $\frac{x}{2L} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$.
Rearranging for $x$: $x^3 = \frac{2q^2 L}{4\pi \varepsilon_0 mg} = \frac{q^2 L}{2\pi \varepsilon_0 mg}$.
Therefore,$x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{1/3}}$.

(B) For each sphere,three forces are acting: $(1)$ Tension force $(T)$,$(2)$ Gravitational force $(Mg)$,and $(3)$ Electrostatic repulsive force $(F)$.
For the first sphere at equilibrium:
$T_1 \cos \theta_1 = M_1 g$ and $T_1 \sin \theta_1 = F$
Dividing these,we get: $\tan \theta_1 = \frac{F}{M_1 g}$
For the second sphere at equilibrium:
$T_2 \cos \theta_2 = M_2 g$ and $T_2 \sin \theta_2 = F$
Dividing these,we get: $\tan \theta_2 = \frac{F}{M_2 g}$
Since the electrostatic force $F$ between the two spheres is the same (by Newton's third law),for $\theta_1 = \theta_2$,we must have $\tan \theta_1 = \tan \theta_2$.
Therefore,$\frac{F}{M_1 g} = \frac{F}{M_2 g}$,which implies $M_1 = M_2$.

(A) The net force on the charge $q_0$ at a distance $y$ from the origin along the $y$-axis is due to the two charges $q$ at $(-a, 0)$ and $(a, 0)$.
By symmetry,the horizontal components of the forces cancel out,and the vertical components add up.
The force from each charge $q$ on $q_0$ is $F = \frac{k q q_0}{r^2}$,where $r^2 = a^2 + y^2$.
The net force is $F_{net} = 2F \sin\theta$,where $\sin\theta = \frac{y}{r} = \frac{y}{\sqrt{a^2 + y^2}}$.
Substituting the values,we get $F_{net} = 2 \left[ \frac{k q q_0}{a^2 + y^2} \right] \cdot \frac{y}{(a^2 + y^2)^{1/2}} = \frac{2 k q q_0 y}{(a^2 + y^2)^{3/2}}$.
For a small displacement $(y << a)$,we can approximate $a^2 + y^2 \approx a^2$.
Thus,$F_{net} \approx \frac{2 k q q_0 y}{a^3}$.
Since $k, q, q_0,$ and $a$ are constants,the net force is proportional to $y$.

(C) Consider a small element of the rod of length $dx$ at a distance $x$ from the point charge $q$,as shown in the figure. Treat this element as a point charge.
The force $dF$ between the charge $q$ and the charge element $dQ$ is given by Coulomb's law:
$dF = \frac{1}{4\pi \epsilon_0} \frac{q \cdot dQ}{x^2}$
Since the charge $Q$ is uniformly distributed over length $L$,the charge per unit length is $\lambda = \frac{Q}{L}$. Therefore,$dQ = \lambda dx = \frac{Q}{L} dx$.
Substituting $dQ$ into the force equation:
$dF = \frac{1}{4\pi \epsilon_0} \frac{q (Q/L) dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \frac{dx}{x^2}$
To find the total force $F$,integrate $dF$ from $x = d$ to $x = d + L$:
$F = \int_{d}^{d+L} \frac{qQ}{4\pi \epsilon_0 L} \frac{dx}{x^2} = \frac{qQ}{4\pi \epsilon_0 L} \left[ -\frac{1}{x} \right]_{d}^{d+L}$
$F = \frac{qQ}{4\pi \epsilon_0 L} \left( -\frac{1}{d+L} - (-\frac{1}{d}) \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{1}{d} - \frac{1}{d+L} \right)$
$F = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{(d+L) - d}{d(d+L)} \right) = \frac{qQ}{4\pi \epsilon_0 L} \left( \frac{L}{d(d+L)} \right)$
$F = \frac{1}{4\pi \epsilon_0} \frac{qQ}{d(d+L)}$

(A) The charges are $q_A = +100 \ \mu C$,$q_B = +100 \ \mu C$,and $q_C = -100 \ \mu C$. The side length $r = 10 \ cm = 0.1 \ m$.
The force on $B$ due to $A$ $(F_A)$ is repulsive and acts along the line $AB$ away from $A$: $F_A = \frac{k |q_A q_B|}{r^2} = \frac{9 \times 10^9 \times (100 \times 10^{-6}) \times (100 \times 10^{-6})}{(0.1)^2} = 9 \times 10^3 \ N$.
The force on $B$ due to $C$ $(F_C)$ is attractive and acts along the line $BC$ towards $C$: $F_C = \frac{k |q_C q_B|}{r^2} = \frac{9 \times 10^9 \times (100 \times 10^{-6}) \times (100 \times 10^{-6})}{(0.1)^2} = 9 \times 10^3 \ N$.
The angle between $F_A$ and $F_C$ is $120^\circ$ because the angle between the lines $AB$ and $BC$ is $60^\circ$,and the force $F_C$ is directed towards $C$ (making an angle of $180^\circ - 60^\circ = 120^\circ$ with $F_A$).
The resultant force $F_{net} = \sqrt{F_A^2 + F_C^2 + 2 F_A F_C \cos(120^\circ)}$.
Since $F_A = F_C = F = 9 \times 10^3 \ N$,$F_{net} = \sqrt{F^2 + F^2 + 2F^2(-0.5)} = \sqrt{F^2} = F = 9 \times 10^3 \ N$.

(C) Let the charges be placed at the corners of a square of side $a$. The charge $q$ at $(a, 0)$ experiences forces from the other three charges.
Let the charges at $(0, 0)$ and $(a, a)$ be $Q$,and the charge at $(0, a)$ be $q$.
The force on $q$ at $(a, 0)$ due to the two $Q$ charges is $F_Q = \sqrt{2} \frac{kQq}{a^2}$ directed towards the origin.
The force on $q$ at $(a, 0)$ due to the charge $q$ at $(0, a)$ is $F_q = \frac{kq^2}{(\sqrt{2}a)^2} = \frac{kq^2}{2a^2}$ directed away from $(0, a)$.
For the net force to be zero,these two forces must be equal in magnitude and opposite in direction:
$\sqrt{2} \frac{kQq}{a^2} + \frac{kq^2}{2a^2} = 0$
$\sqrt{2} Q + \frac{q}{2} = 0$
$q = -2\sqrt{2} Q$.

(A) Initially,the charges are $q_1 = +7 \ \mu C$ and $q_2 = -5 \ \mu C$. The force between them is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2} = k \frac{|(7 \times 10^{-6})(-5 \times 10^{-6})|}{r^2} = k \frac{35 \times 10^{-12}}{r^2}$.
After adding $-2 \ \mu C$ to both,the new charges are:
$q_1' = 7 \ \mu C - 2 \ \mu C = +5 \ \mu C$
$q_2' = -5 \ \mu C - 2 \ \mu C = -7 \ \mu C$.
The new force $F'$ is:
$F' = k \frac{|q_1' q_2'|}{r^2} = k \frac{|(5 \times 10^{-6})(-7 \times 10^{-6})|}{r^2} = k \frac{35 \times 10^{-12}}{r^2}$.
Comparing the two expressions,we see that $F' = F$.

(C) Let the two identical charges be $Q$ and the third charge be $q$. The distance between the two charges is $d$. The distance of the third charge from the midpoint of the line joining the two charges is $x$.
The force exerted by each charge $Q$ on $q$ is $F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{x^2 + (d/2)^2}$.
The resultant force $F_{res}$ on $q$ along the perpendicular bisector is $F_{res} = 2F \cos\theta$,where $\cos\theta = \frac{x}{\sqrt{x^2 + (d/2)^2}}$.
Substituting $F$ and $\cos\theta$,we get $F_{res} = \frac{2Qqx}{4\pi\epsilon_0 (x^2 + d^2/4)^{3/2}}$.
To find the maximum force,we differentiate $F_{res}$ with respect to $x$ and set it to zero: $\frac{dF_{res}}{dx} = 0$.
Using the quotient rule or product rule,we get $(x^2 + d^2/4)^{3/2} - x \cdot \frac{3}{2}(x^2 + d^2/4)^{1/2} \cdot 2x = 0$.
$(x^2 + d^2/4) - 3x^2 = 0 \implies 2x^2 = d^2/4 \implies x^2 = d^2/8$.
Therefore,$x = \frac{d}{\sqrt{8}} = \frac{d}{2\sqrt{2}}$.

(A) According to Coulomb's Law,the initial force $F$ is given by: $F = k \frac{q_1 q_2}{r^2} = k \frac{(3 \times 10^{-6})(8 \times 10^{-6})}{r^2} = 40 \ N$.
After adding $-5 \ \mu C$ to both charges,the new charges are: $q_1' = 3 - 5 = -2 \ \mu C$ and $q_2' = 8 - 5 = 3 \ \mu C$.
The new force $F'$ is: $F' = k \frac{q_1' q_2'}{r^2} = k \frac{(-2 \times 10^{-6})(3 \times 10^{-6})}{r^2}$.
Taking the ratio: $\frac{F'}{F} = \frac{(-2)(3)}{(3)(8)} = \frac{-6}{24} = -\frac{1}{4}$.
Therefore,$F' = -\frac{1}{4} \times 40 = -10 \ N$.

(D) The electrostatic force between two charges $Q_1$ and $Q_2$ separated by distance $R$ is given by $F = k \frac{Q_1 Q_2}{R^2}$.
Since $Q_1 + Q_2 = Q$,we can write $Q_2 = Q - Q_1$.
Substituting this into the force equation: $F = \frac{k}{R^2} (Q_1 Q - Q_1^2)$.
To find the maximum force,we differentiate $F$ with respect to $Q_1$ and set it to zero: $\frac{dF}{dQ_1} = \frac{k}{R^2} (Q - 2Q_1) = 0$.
Solving for $Q_1$,we get $Q - 2Q_1 = 0$,which implies $Q_1 = \frac{Q}{2}$.
Since $Q_2 = Q - Q_1$,we find $Q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$.
Thus,the force is maximum when $Q_1 = Q_2 = \frac{Q}{2}$.

(A) Initially,the charges are $+Q$ and $-Q$ separated by distance $r$. The electrostatic force is $F = k\frac{|(+Q)(-Q)|}{r^2} = k\frac{Q^2}{r^2}$.
When $75\%$ of the charge from the first sphere $(+Q)$ is transferred to the second sphere $(-Q)$,the new charge on the first sphere is $Q' = Q - 0.75Q = 0.25Q = \frac{Q}{4}$.
The new charge on the second sphere is $Q'' = -Q + 0.75Q = -0.25Q = -\frac{Q}{4}$.
The new force $F'$ is $F' = k\frac{|(Q/4)(-Q/4)|}{r^2} = k\frac{Q^2/16}{r^2} = \frac{1}{16} \left( k\frac{Q^2}{r^2} \right) = \frac{F}{16}$.