Electric Charge, It's Properties and Method of Charging Questions in English

(A) The Faraday is a unit of electric charge used in physics and chemistry.
$1$ Faraday is defined as the magnitude of the total electric charge per mole of electrons.
Its value is approximately $96485 \ C/mol$,often rounded to $96500 \ C/mol$.
Therefore,Faraday is the unit of charge.

(A) The fundamental unit of charge is the electron,which carries a negative charge.
When a neutral body gains excess electrons,the total number of electrons becomes greater than the number of protons.
Since protons are positively charged and electrons are negatively charged,an excess of electrons results in a net negative charge on the body.
Therefore,giving excess electrons to a body makes it negatively charged.

(C) According to the principle of quantization of electric charge,the electric charge of any object is always an integral multiple of the elementary charge $(e)$.
The elementary charge is the charge of an electron or a proton,which is $e = 1.6 \times 10^{-19} \text{ C}$.
Therefore,the minimum possible charge on any free object is $1.6 \times 10^{-19} \text{ C}$.

(A) When a charge $ + Q$ is given to an isolated solid metallic sphere,the charges experience mutual electrostatic repulsion.
Because the material is a conductor,the charges are free to move.
To minimize the potential energy of the system,the charges redistribute themselves such that they reside entirely on the outer surface of the sphere.
Due to the spherical symmetry of the conductor,the charge distribution on the surface is uniform.

(C) The correct answer is $C$.
For a metallic sphere, whether it is solid or hollow, the excess charge resides entirely on its outer surface due to the electrostatic repulsion between like charges.
Since both spheres have the same radius $R$, they have the same surface area $A = 4\pi R^2$.
Therefore, both spheres can hold the same maximum amount of charge before the electric field at the surface exceeds the dielectric strength of the surrounding medium (causing ionization/discharge).

(C) The correct answer is $C$. When a carrier moves,friction with the air causes static electricity to build up on its surface. If this charge accumulates,it could produce a spark,which is dangerous for inflammable materials. Metallic ropes are used to provide a continuous conducting path to the earth,allowing the induced charge to discharge safely into the ground.

(A) The charge $Q$ on a body is given by the quantization of charge formula: $Q = ne$.
Here,$n = 10^{14}$ is the number of electrons removed,and $e = 1.6 \times 10^{-19} \ C$ is the elementary charge.
Since electrons are removed from the neutral sphere,the sphere becomes positively charged.
$Q = 10^{14} \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-5} \ C$.
To convert this into microcoulombs $(\mu C)$,we multiply by $10^6$:
$Q = 1.6 \times 10^{-5} \times 10^6 \ \mu C = 1.6 \times 10^1 \ \mu C = 16 \ \mu C$.

(D) The quantization of charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge.
Given $Q = 14.4 \times 10^{-19} \ C$ and $e = 1.6 \times 10^{-19} \ C$.
We need to find $n = \frac{Q}{e}$.
$n = \frac{14.4 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{14.4}{1.6} = 9$.
Since the conductor has a positive charge,it indicates a deficiency of electrons.
Therefore,the conductor has $9$ electrons in short.

(B) According to the quantization of charge,the total charge $q$ is given by $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given,$q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
Therefore,$n = \frac{q}{e} = \frac{1}{1.6 \times 10^{-19}}$.
$n = \frac{1}{1.6} \times 10^{19} = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Thus,the number of electrons in one coulomb of charge is $6.25 \times 10^{18}$.

(D) When a body is given a positive charge,it loses electrons. Since electrons have a finite mass $(m_e \approx 9.1 \times 10^{-31} \ kg)$,the mass of sphere $A$ decreases.
When a body is given a negative charge,it gains electrons. Since electrons have a finite mass,the mass of sphere $B$ increases.
Therefore,the mass of sphere $B$ becomes greater than the mass of sphere $A$.

(B) The quantization of electric charge is given by the formula $Q = ne$,where $n$ is the number of electrons removed or added and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \, C)$.
Since electrons are removed from the neutral metal plate,the plate becomes positively charged.
Given $n = 10^{19}$ and $e = 1.6 \times 10^{-19} \, C$,the charge $Q$ is:
$Q = (+10^{19}) \times (1.6 \times 10^{-19} \, C)$
$Q = 1.6 \times 10^{19-19} \, C$
$Q = 1.6 \times 10^0 \, C$
$Q = +1.6 \, C$.

(C) An $\alpha$-particle is a helium nucleus,which consists of $2$ protons and $2$ neutrons.
Since neutrons are electrically neutral,the total charge on the $\alpha$-particle is equal to the charge of $2$ protons.
The charge of a single proton is $e = 1.6 \times 10^{-19} \, C$.
Therefore,the total charge $Q$ is given by $Q = 2e = 2 \times (1.6 \times 10^{-19} \, C) = 3.2 \times 10^{-19} \, C$.

(B) When a glass rod is rubbed with silk,electrons are transferred from the glass rod to the silk due to the difference in their work functions.
Since the glass rod loses electrons,it acquires a positive charge.
Since the silk gains electrons,it acquires a negative charge.
Therefore,the glass rod gives electrons to the silk.

(C) The quantization of charge is given by the formula $Q = Ne$,where $Q$ is the total charge,$N$ is the number of electrons,and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
Given: $Q = 80 \ \mu C = 80 \times 10^{-6} \ C$.
Using the formula $N = \frac{Q}{e}$:
$N = \frac{80 \times 10^{-6}}{1.6 \times 10^{-19}}$
$N = \frac{80}{1.6} \times 10^{-6 - (-19)}$
$N = 50 \times 10^{13} = 5 \times 10^{14}$.
Therefore,the number of additional electrons is $5 \times 10^{14}$.

(C) Given that $(2, 3)$ and $(4, 5)$ show repulsion,balls $2$ and $3$ have the same charge,and balls $4$ and $5$ have the same charge.
Since $(1, 2)$,$(2, 4)$,and $(4, 1)$ show attraction,if $1$ were charged,it would have to have a charge opposite to $2$ and $4$. However,if $2$ and $4$ have opposite charges,$1$ cannot simultaneously attract both unless $1$ is neutral.
When a neutral object is brought near a charged object,it experiences electrostatic induction,resulting in an attractive force regardless of the sign of the charge on the other object.
Thus,ball $1$ must be neutral.

(B) When a body is connected to the earth,the potential of the body becomes equal to the potential of the earth (which is $0 \ V$).
If electrons flow from the earth into the body,it implies that the body must have been at a higher potential than the earth,which is only possible if the body was positively charged.
The electrons flow to neutralize the positive charge on the body.
Therefore,the body was initially charged positively.

(A) According to the quantization of charge,the total charge $q$ is given by $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given: $q = 1.6 \, C$ and $e = 1.6 \times 10^{-19} \, C$.
To find the number of electrons $n$,we use the formula $n = \frac{q}{e}$.
Substituting the values: $n = \frac{1.6}{1.6 \times 10^{-19}} = \frac{1}{10^{-19}} = 10^{19}$.
Therefore,the number of electrons is $10^{19}$.

(A) The ability of a conductor to retain charge depends on the uniformity of the surface charge distribution.
For a spherical conductor,the charge distributes itself uniformly over the entire surface due to symmetry.
In non-spherical conductors (like pear-shaped or lightning conductors),the charge density is higher at regions with smaller radii of curvature (sharp points).
This leads to a higher electric field intensity at sharp points,which causes the ionization of surrounding air,resulting in the leakage of charge (corona discharge).
Therefore,a spherical conductor is best suited to retain charges for a longer time as it minimizes charge leakage.

(B) The density of water is $1\,g/cc$,so the mass of $500\,cc$ of water is $500\,g$.
The molar mass of water $(H_2O)$ is $18\,g/mol$.
The number of moles of water is $n_{moles} = \frac{500}{18} \approx 27.78\,mol$.
Each molecule of water $(H_2O)$ contains $10$ protons ($1$ from each $H$ atom and $8$ from the $O$ atom).
The total number of protons $N_p$ is given by $N_p = n_{moles} \times N_A \times 10$,where $N_A = 6.022 \times 10^{23}\,mol^{-1}$.
$N_p = \frac{500}{18} \times 6.022 \times 10^{23} \times 10 \approx 1.673 \times 10^{26}$.
The total charge $Q$ is $N_p \times e$,where $e = 1.6 \times 10^{-19}\,C$.
$Q = 1.673 \times 10^{26} \times 1.6 \times 10^{-19} \approx 2.67 \times 10^7\,C$.

(D) In Millikan's oil drop experiment,the oil used must be non-volatile. If the oil were volatile,the drops would evaporate during the experiment,causing their mass and radius to change continuously. This would make it impossible to accurately determine the charge on the electron. Therefore,a non-volatile oil is chosen so that the size and mass of the droplets remain constant throughout the observation.

(A) The quantization of charge states that any charge $q$ is an integer multiple of the elementary charge $e$,given by $q = ne$,where $n$ is an integer.
To find $e$,we look for the greatest common divisor of the observed charges.
Dividing each charge by the smallest value $(6.563 \times 10^{-19} \ C)$:
$1 : 1.25 : 1.75 : 2.0 : 2.5 : 2.75$
Multiplying by $4$ to get integers:
$4 : 5 : 7 : 8 : 10 : 11$
These represent the values of $n$ for each charge.
We calculate $e$ by taking the average: $e = \frac{\sum q_i}{\sum n_i} = \frac{(6.563 + 8.204 + 11.50 + 13.13 + 16.48 + 18.09) \times 10^{-19}}{4 + 5 + 7 + 8 + 10 + 11}$
$e = \frac{73.967 \times 10^{-19}}{45} \approx 1.641 \times 10^{-19} \ C$.

(D) The quantization of electric charge,which states that any electric charge $q$ is an integral multiple of the fundamental electronic charge $e$ (i.e.,$q = ne$,where $n$ is an integer),was experimentally demonstrated by Robert $A$. Millikan in his famous Oil Drop Experiment. Therefore,the correct option is $D$.

(C) The charge on an electron was determined by Robert Millikan in $1906$ through his famous oil drop experiment.
He measured the charge of individual oil droplets and found that the charge was always an integer multiple of a fundamental value,which he identified as the charge of an electron.

(B) In Millikan's oil drop experiment,the charge $q$ on any oil drop is always an integral multiple of the elementary charge $e$. This is expressed as $q = ne$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
Given $e = 1.6 \times 10^{-19} \text{ C}$.
Option $B$ contains $2e$ (which is $2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ C}$) and $10e$ (which is $10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18} \text{ C}$).
Since both $2$ and $10$ are integers,these charges are possible on the oil drops.

(C) The kinetic energy $(K)$ gained by a charged particle accelerated through a potential difference $(V)$ is given by the formula: $K = qV$.
For a proton,the charge is $q = +e$.
Given the potential difference $V = 1 \, V$.
Substituting these values,we get: $K = (1e) \times (1 \, V) = 1 \, eV$.
Therefore,the kinetic energy of the proton is $1 \, eV$.

(B) According to the quantization of electric charge,the total charge $Q$ on an object is given by $Q = ne$,where $n$ is an integer representing the number of excess electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
Given $Q = 6.35 \times 10^{-19} \ C$.
We need to find $n = \frac{Q}{e}$.
Substituting the values: $n = \frac{6.35 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{6.35}{1.6} \approx 3.968$.
Since the number of electrons $n$ must be an integer,we round $3.968$ to the nearest integer,which is $4$.
Therefore,the number of excess electrons on the drop is $4$.

(D) Ionic bonds are formed when atoms with low ionization energies,which readily lose electrons,interact with atoms that tend to acquire excess electrons.
In such a crystal,the former atoms transfer electrons to the latter,resulting in the formation of positive and negative ions.
These ions are then held together by strong electrostatic forces of attraction,which is characteristic of ionic bonding.
Therefore,a crystal structure consisting of alternate and evenly spaced positive and negative ions is ionic in nature.

(B) According to the quantization of charge,the total charge $Q$ is given by $Q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given $Q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
We need to find $n = Q / e$.
$n = 1 / (1.6 \times 10^{-19})$
$n = 10^{19} / 1.6$
$n = 0.625 \times 10^{19}$
$n = 6.25 \times 10^{18}$ electrons.

(C) An $\alpha$-particle is a helium nucleus,which consists of $2$ protons and $2$ neutrons.
Since neutrons are electrically neutral,the total charge on an $\alpha$-particle is equal to the charge of $2$ protons.
The charge of a single proton is $e = 1.6 \times 10^{-19} \, C$.
Therefore,the charge of an $\alpha$-particle is $q = 2e = 2 \times (1.6 \times 10^{-19} \, C) = 3.2 \times 10^{-19} \, C$.

(B) static electric charge is a charge at rest.
$1$. It satisfies the conservation of electric charge.
$2$. It produces an electric field in the space around it.
$3$. Electric charge is quantized $(q = ne)$.
$4$. $A$ magnetic field is produced only by a moving electric charge (current).
Therefore,a static electric charge does not generate a magnetic field.

(A) body becomes negatively charged when it gains an excess of electrons. Since electrons carry a negative charge,adding them to a neutral body increases the total negative charge,making the body negatively charged. Removing electrons would make it positively charged,and protons or neutrons cannot be easily transferred in standard electrostatic charging processes.

(A) The quantization of electric charge is given by the formula $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given: $n = 10^{14}$ electrons.
Since electrons are removed,the sphere becomes positively charged.
The elementary charge $e = 1.6 \times 10^{-19} \ C$.
Therefore,$q = 10^{14} \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-5} \ C$.
To convert this into $\mu C$ (microcoulombs),we multiply by $10^6$.
$q = 1.6 \times 10^{-5} \times 10^6 \ \mu C = 1.6 \times 10^1 \ \mu C = 16 \ \mu C$.

(B) The quantization of electric charge was experimentally demonstrated by Robert $A$. Millikan in his famous oil drop experiment.
In this experiment,tiny oil droplets were sprayed into a chamber.
By observing the motion of these droplets under the influence of gravity and a uniform electric field,Millikan found that the charge $q$ on any oil drop was always an integer multiple of a fundamental charge $e$,i.e.,$q = ne$,where $n = 1, 2, 3, ...$.
This proved that electric charge is quantized.

(C) The net electric charge of a conductor remains constant when it is brought near a charged body due to the phenomenon of electrostatic induction.
In electrostatic induction,the charges within the conductor redistribute themselves (positive charges move to one side and negative charges to the other),but the total net charge of the conductor remains zero (if it was initially neutral) or unchanged (if it was initially charged).
Options $A$,$B$,and $D$ involve the transfer or addition/removal of charge,which changes the net charge of the conductor.

(D) According to the quantization of electric charge,the total charge $q$ on any body is always an integral multiple of the elementary charge $e$,where $e = 1.6 \times 10^{-19} \ C$.
The formula is $q = n \times e$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
To check if a charge is possible,we calculate $n = q / e$:
For option $A$: $n = (4.0 \times 10^{-19}) / (1.6 \times 10^{-19}) = 2.5$ (Not an integer).
For option $B$: $n = (6.0 \times 10^{-19}) / (1.6 \times 10^{-19}) = 3.75$ (Not an integer).
For option $C$: $n = (10.0 \times 10^{-19}) / (1.6 \times 10^{-19}) = 6.25$ (Not an integer).
Since $n$ is not an integer for any of the given options,none of these charges are possible in Millikan's oil drop experiment. Therefore,the correct answer is $D$.

(C) An $\alpha$-particle is a helium nucleus,which consists of $2$ protons and $2$ neutrons.
Since the charge of a proton is $e = 1.6 \times 10^{-19} \, C$ and neutrons are neutral,the total charge $Q$ of an $\alpha$-particle is given by $Q = +2e$.
Substituting the value of $e$:
$Q = 2 \times (1.6 \times 10^{-19} \, C) = 3.2 \times 10^{-19} \, C$.

(B) Millikan's oil drop experiment demonstrated that electric charge is quantized.
This means that the total charge $q$ on any object is always an integer multiple of the elementary charge $e$ (where $e \approx 1.6 \times 10^{-19} \, C$).
Mathematically, $q = n \cdot e$, where $n$ is an integer $(n = 1, 2, 3, \dots)$.
Looking at the options, $1.6 \times 10^{-19} \, C$ corresponds to $n=1$ $(1e)$ and $2e$ corresponds to $n=2$.
Since $2.5e$ and $1.5e$ are not integer multiples of $e$, they cannot exist on the oil drop.
Therefore, the set of charges $2e$ and $1.6 \times 10^{-19} \, C$ (which is $1e$) are valid values.

(D) An insulating material does not allow charge to flow freely through it. When a charge $+Q$ is given to an insulating solid sphere,the charge remains where it is placed and does not redistribute itself to the surface as it would in a conductor. Therefore,the charge distribution will be non-uniform throughout the volume of the sphere.

(D) The quantization of electric charge states that the total charge $q$ of a body is given by $q = ne$,where $n$ is an integer and $e$ is the elementary charge.
This fundamental property,which confirms that electric charge is an integral multiple of the electronic charge,was experimentally demonstrated by Robert $A$. Millikan through his famous oil-drop experiment.

(A) The number of $\alpha$-particles falling on the sphere in $t$ seconds is $N = 10^{12}t$.
Each $\alpha$-particle carries a charge of $q_{\alpha} = +2e$,where $e = 1.6 \times 10^{-19} \, C$.
The total charge $Q$ accumulated in time $t$ is given by $Q = N \times q_{\alpha} = (10^{12}t) \times (2e)$.
Given $Q = 2 \, \mu C = 2 \times 10^{-6} \, C$.
Equating the values: $2 \times 10^{-6} = 10^{12} \times t \times 2 \times 1.6 \times 10^{-19}$.
$2 \times 10^{-6} = 3.2 \times 10^{-7} \times t$.
$t = \frac{2 \times 10^{-6}}{3.2 \times 10^{-7}} = \frac{20}{3.2} = 6.25 \, s$.

(D) Let the initial charge on each sphere $B$ and $C$ be $q$ and the distance between them be $r$. The initial force is $F = \frac{kq^2}{r^2}$.
When uncharged sphere $A$ is brought in contact with $B$,the charge on $B$ becomes $q/2$ and on $A$ becomes $q/2$.
Now,when sphere $A$ (having charge $q/2$) is brought in contact with $C$ (having charge $q$),the total charge is $q + q/2 = 3q/2$. Since the spheres are identical,the charge is shared equally,so each gets $(3q/2) / 2 = 3q/4$.
Now,the charge on $B$ is $q/2$ and the charge on $C$ is $3q/4$.
The new force $F' = \frac{k(q/2)(3q/4)}{r^2} = \frac{3kq^2}{8r^2}$.
Since $F = \frac{kq^2}{r^2}$,we get $F' = \frac{3}{8}F$.

(C) The charge on the polythene is $Q = -2 \times 10^{-7} \ C$. Since the charge is negative,electrons are transferred from the wool to the polythene.
Using the quantization of charge,$Q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron.
The number of electrons transferred is $n = \frac{|Q|}{e} = \frac{2 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.25 \times 10^{12}$.
The mass of one electron is $m_e = 9.1 \times 10^{-31} \ kg$.
The total mass transferred to the polythene is $M = n \times m_e$.
$M = (1.25 \times 10^{12}) \times (9.1 \times 10^{-31}) \ kg$.
$M = 11.375 \times 10^{-19} \ kg \approx 11.38 \times 10^{-19} \ kg$.

(C) According to the quantization of electric charge,the total charge $q$ on a body is given by $q = ne$,where $n$ is an integer $(n = 1, 2, 3, \dots)$ and $e = 1.6 \times 10^{-19} \ C$ is the elementary charge.
For $q = 1.6 \times 10^{-18} \ C$,$n = q/e = (1.6 \times 10^{-18}) / (1.6 \times 10^{-19}) = 10$,which is an integer.
For $q = 1.6 \times 10^{-19} \ C$,$n = q/e = 1$,which is an integer.
For $q = 1.6 \times 10^{-20} \ C$,$n = q/e = (1.6 \times 10^{-20}) / (1.6 \times 10^{-19}) = 0.1$,which is not an integer.
Since $n$ must be an integer,a charge of $1.6 \times 10^{-20} \ C$ is not possible.

(B) According to the quantization of charge,the total charge $Q$ is given by $Q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given $Q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula for $n$: $n = \frac{Q}{e}$.
Substituting the values: $n = \frac{1}{1.6 \times 10^{-19}} = \frac{10^{19}}{1.6} = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,the number of electrons in $1 \ C$ of charge is $6.25 \times 10^{18}$.

(A) The total number of atoms is $N = 5 \times 10^{21}$.
The number of atoms from which one electron is removed is $n = 0.01\% \text{ of } N$.
$n = \frac{0.01}{100} \times 5 \times 10^{21} = 10^{-4} \times 5 \times 10^{21} = 5 \times 10^{17}$.
Since one electron is removed from each of these atoms,the total number of electrons removed is $n = 5 \times 10^{17}$.
The charge acquired by the sphere due to the removal of electrons is $Q = +ne$.
$Q = (5 \times 10^{17}) \times (1.6 \times 10^{-19} \ C)$.
$Q = 8 \times 10^{-2} \ C = 0.08 \ C$.
Since electrons are removed,the sphere becomes positively charged,so $Q = +0.08 \ C$.

(D) $(1)$ Charging by conduction depends on the geometry and size of the bodies,not just half of the initial charge.
$(2)$ Charge is invariant with velocity; it does not change as speed increases.
$(3)$ Charge cannot exist without matter (mass).
$(4)$ Repulsion is the true test of electrification because attraction can occur between a charged and an uncharged body due to induction,whereas repulsion only occurs between two like-charged bodies.

(C) When a body is charged by rubbing,electrons are transferred from one body to another.
If a body gains electrons,it becomes negatively charged and its mass increases slightly,leading to a slight increase in weight.
If a body loses electrons,it becomes positively charged and its mass decreases slightly,leading to a slight decrease in weight.
Therefore,the weight may increase or decrease depending on whether the body is charged negatively or positively.

(D) Since the spheres are identical,when two spheres are touched,the total charge is shared equally between them.
Initial state: $Q_A = Q, Q_B = 0, Q_C = 0$.
$(i)$ $A$ and $B$ touch: $Q_A = Q_B = \frac{Q+0}{2} = \frac{Q}{2}$. Charges: $(\frac{Q}{2}, \frac{Q}{2}, 0)$.
$(ii)$ $B$ and $C$ touch: $Q_B = Q_C = \frac{Q/2 + 0}{2} = \frac{Q}{4}$. Charges: $(\frac{Q}{2}, \frac{Q}{4}, \frac{Q}{4})$.
$(iii)$ $C$ and $A$ touch: $Q_C = Q_A = \frac{Q/4 + Q/2}{2} = \frac{3Q/4}{2} = \frac{3Q}{8}$. Charges: $(\frac{3Q}{8}, \frac{Q}{4}, \frac{3Q}{8})$.
$(iv)$ $A$ and $B$ touch: $Q_A = Q_B = \frac{3Q/8 + Q/4}{2} = \frac{5Q/8}{2} = \frac{5Q}{16}$. Charges: $(\frac{5Q}{16}, \frac{5Q}{16}, \frac{3Q}{8})$.
$(v)$ $B$ and $C$ touch: $Q_B = Q_C = \frac{5Q/16 + 3Q/8}{2} = \frac{11Q/16}{2} = \frac{11Q}{32}$. Charges: $(\frac{5Q}{16}, \frac{11Q}{32}, \frac{11Q}{32})$.

(A) When two objects are rubbed against each other,electrons are transferred from one object to the other due to the difference in their electron affinity.
One object loses electrons and becomes positively charged,while the other gains electrons and becomes negatively charged.
Since the two objects now carry opposite charges,they will exert an attractive force on each other when placed at a distance.
Therefore,the nature of the electric force is attractive.

(C) The charge $q$ is given by $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(1.6 \times 10^{-19} \, C)$.
Since the body is given a positive charge of $+2 \, C$,it means electrons are removed from the body.
The number of electrons removed is $n = \frac{q}{e} = \frac{2}{1.6 \times 10^{-19}} = 1.25 \times 10^{19}$.
The change in mass $\Delta m$ is equal to the mass of the removed electrons: $\Delta m = n \times m_e$,where $m_e = 9.1 \times 10^{-31} \, kg$.
$\Delta m = (1.25 \times 10^{19}) \times (9.1 \times 10^{-31} \, kg) = 11.375 \times 10^{-12} \, kg$.
Therefore,the change in mass is approximately $11.37 \times 10^{-12} \, kg$.