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

(N/A) Quantization of electric charge is the property by virtue of which all free charges are integral multiples of a basic unit of charge denoted by $e$. Thus,charge $q$ on a body is always given by $q = n \cdot e$,where $n$ is an integer $(n = 0, \pm 1, \pm 2, \dots)$ and $e \approx 1.602 \times 10^{-19} \ C$. The $SI$ unit of electric charge is the Coulomb $(C)$.

(N/A) The fundamental electric charge is the magnitude of the charge of a single electron or proton.
Its $SI$ unit is the Coulomb,denoted by the symbol $C$.
The value of the fundamental electric charge is approximately $e = 1.602 \times 10^{-19} \ C$.

(A) The quantization of electric charge was first experimentally demonstrated by $Robert \ Millikan$ in $1912$ through his famous $Oil \ Drop \ Experiment$.
In this experiment, he measured the electric charge on tiny oil droplets suspended in an electric field.
He observed that the charge on each droplet was always an integer multiple of a fundamental unit of charge, $e \approx 1.602 \times 10^{-19} \ C$.
This provided direct evidence that electric charge is quantized, meaning it exists in discrete packets rather than a continuous flow.

(N/A) The $Coulomb$ $(C)$ is the $SI$ unit of electric charge.
One $Coulomb$ is defined as the quantity of charge that passes through a cross-section of a conductor in $1 \ s$ when a steady current of $1 \ A$ flows through it.
Mathematically,it is expressed as $Q = I \times t$,where $I = 1 \ A$ and $t = 1 \ s$,resulting in $Q = 1 \ C$.

(A) The quantization of charge states that charge $q$ is always an integer multiple of the elementary charge $e$, i.e., $q = ne$, where $n$ is an integer.
At the macroscopic level, the charges involved are extremely large (e.g., $1 \mu C$ contains approximately $6.25 \times 10^{12}$ electrons).
Because the number of charges $n$ is so large, the discrete nature of charge becomes negligible, and charge can be treated as a continuous quantity.
Therefore, we can neglect the quantization of charge at the macroscopic level.

(B) When the linear size of charged bodies is much smaller than the distance separating them,the size of the bodies may be ignored,and the charged bodies are treated as point charges.

(A) The magnetic field $(B)$ is produced by moving charges or currents,which are vector sources.
The electric field $(E)$ is produced by stationary charges,which act as scalar sources.
Therefore,the magnetic field vector component is due to moving charges (or current),and the electric field scalar component is due to stationary charges.

(B) The electron is a subatomic particle that carries a negative elementary electric charge.
According to the Millikan's oil drop experiment,the magnitude of the charge on an electron is approximately $1.602 \times 10^{-19} \ C$.
Since the electron is negatively charged,its charge is represented as $-1.6 \times 10^{-19} \ C$.

(B) In a neutral atom, the number of electrons is equal to the number of protons.
Each electron carries a charge of $-e$ (where $e = 1.6 \times 10^{-19} \ C$).
Each proton carries a charge of $+e$.
Since the number of electrons $(Z)$ equals the number of protons $(Z)$, the total charge of the electrons is $-Ze$, and the total charge of the protons is $+Ze$.
The net charge of the atom is $(-Ze) + (+Ze) = 0$.
However, the question asks for the total charge of the electrons specifically, which is $-Ze$ (where $Z$ is the atomic number).

(N/A) When a comb is run through dry hair,it acquires an electric charge due to friction. The pieces of paper are initially neutral. When the charged comb is brought near the paper,it induces a redistribution of charges in the paper,a process called electrostatic induction. This creates a dipole moment in the paper. Since the electric field produced by the comb is non-uniform,the force on the induced charges is not balanced,resulting in a net attractive force towards the comb.

(C) Coulomb's law is fundamentally defined for point charges.
When we have a system of multiple point charges (discrete charge distribution),the total electric field at a point is the vector sum of the individual electric fields produced by each point charge,which is known as the superposition principle.
For continuous charge distributions,we treat the distribution as a collection of infinitely many infinitesimal point charges and integrate over the region.
Therefore,both discrete and continuous charge distributions can be analyzed using these principles.

(A) The mass of the coin is $W = 0.75\, g$. Assuming the coin is primarily composed of Aluminum $(Al)$ for calculation purposes,the molar mass of $Al$ is $M = 26.98\, g/mol$.
The number of atoms $N$ in the coin is given by $N = \frac{W}{M} \times N_A$,where $N_A = 6.023 \times 10^{23}\, mol^{-1}$.
$N = \frac{0.75}{26.98} \times 6.023 \times 10^{23} \approx 1.674 \times 10^{22}$ atoms.
The atomic number of $Al$ is $Z = 13$,meaning each atom contains $13$ protons and $13$ electrons.
The total charge $Q$ is given by $Q = N \times Z \times e$,where $e = 1.6 \times 10^{-19}\, C$.
$Q = 1.674 \times 10^{22} \times 13 \times 1.6 \times 10^{-19} \approx 3.48 \times 10^{4}\, C$.
Thus,the total amount of positive or negative charge is approximately $3.48 \times 10^{4}\, C$.

(B) An alpha-volt $(\alpha-V)$ is defined as the energy acquired by an $\alpha$-particle when it is accelerated through a potential difference of $1 \,V$.
The charge of an $\alpha$-particle is $q = +2e$,where $e$ is the elementary charge.
The energy $E$ acquired by a charged particle accelerated through a potential difference $V$ is given by $E = qV$.
Substituting the values for an $\alpha$-particle:
$E = (2e) \times (1 \,V) = 2 \,eV$.
Therefore,$1 \alpha-V = 2 \,eV$.

(A) The water molecule $(H_2O)$ is a polar molecule,meaning it has a permanent electric dipole moment.
When a charged rod (whether positively or negatively charged) is brought near a neutral stream of water,it induces a redistribution of charge within the water molecules due to electrostatic induction.
Because the water molecules are polar,the end of the molecule with the opposite charge to the rod is attracted,while the end with the same charge is repelled.
Since the attracted end is closer to the rod than the repelled end,the net force is always attractive.
Therefore,the water stream will bend towards the rod regardless of whether the rod is positively or negatively charged.

(B) The correct answer is $B$.
Electric repulsion occurs only when two bodies possess like charges (both positive or both negative).
In contrast,attraction can occur between two oppositely charged bodies $OR$ between a charged body and an uncharged (neutral) body due to the process of electrostatic induction.
Therefore,attraction is not a definitive test for the presence of charge,whereas repulsion is the only sure test for the presence of electric charge.

(B) Let the charges on objects $P, Q,$ and $R$ be $q_P, q_Q,$ and $q_R$ respectively.
Since $P$ and $Q$ repel each other,they must have charges of the same sign (both positive or both negative). Thus,$q_P$ and $q_Q$ have the same sign.
Since $P$ and $R$ attract each other,they must have charges of opposite signs. Thus,$q_P$ and $q_R$ have opposite signs.
Since $q_Q$ has the same sign as $q_P$,and $q_R$ has the opposite sign to $q_P$,it follows that $q_Q$ and $q_R$ must have opposite signs.
Objects with opposite charges attract each other. Therefore,the force between $Q$ and $R$ is attractive.

(C) The correct statement is $(c)$.
According to the property of quantization of electric charge,the total charge $q$ on any body is always an integral multiple of the elementary charge $e$ (the magnitude of the charge of an electron).
Mathematically,this is expressed as $q = \pm ne$,where $n$ is an integer $(n = 1, 2, 3, ...)$ and $e \approx 1.6 \times 10^{-19} \ C$.
Option $(a)$ is incorrect because charge is conserved,not interconvertible with energy in this manner.
Option $(b)$ is incorrect because electric charge is a relativistic invariant; it does not change with the velocity of the particle.
Option $(d)$ is incorrect because a body can also have a negative charge.

(D) The correct option is $D$.
Electric charge is always associated with mass. $A$ particle cannot possess an electric charge without having a corresponding mass.
Therefore,a particle with zero rest mass (such as a photon) can never have an electric charge. Thus,the statement that a particle having zero rest mass can have non-zero charge is false.

(A) As the vehicle moves,static charge is developed on its body due to friction with the air. This accumulated charge can lead to a spark,which is dangerous for vehicles carrying inflammable materials. Metallic chains are used to provide a conducting path to the ground,allowing the excess charge to flow away safely and preventing the risk of combustion.

(D) neutral conducting sphere contains an equal number of positive and negative charges,resulting in a net charge of $0$.
When a positive point charge is placed outside the sphere,it induces a redistribution of charges within the conductor.
The negative charges are attracted towards the side of the sphere closest to the positive point charge,while the positive charges are repelled to the far side.
However,this process is purely an internal redistribution of existing charges.
Since no charge is added to or removed from the isolated conducting sphere,the total net charge on the sphere remains $0$.

(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 $(n = 0, \pm 1, \pm 2, \dots)$ and $e = 1.6 \times 10^{-19} \ C$.
For a charge to be possible,$n = q/e$ must be an integer.
$(a)$ $n = (3.2 \times 10^{-20}) / (1.6 \times 10^{-19}) = 0.2$ (Not an integer)
$(b)$ $n = (3.2 \times 10^{-18}) / (1.6 \times 10^{-19}) = 20$ (Integer)
$(c)$ $n = (3 \times 10^{-19}) / (1.6 \times 10^{-19}) = 1.875$ (Not an integer)
$(d)$ $n = (6 \times 10^{-17}) / (1.6 \times 10^{-19}) = 375$ (Integer)
Therefore,charges in options $b$ and $d$ are possible.

(A) Concept: When a conductor is charged,the charge resides entirely on its outer surface due to the mutual repulsion between like charges.
Since both spheres have the same radius,their outer surface areas are identical.
Therefore,the maximum amount of charge that can be stored on the surface before dielectric breakdown of the surrounding medium occurs is the same for both.
Thus,both metallic spheres (solid or hollow) can be charged equally.

(D) The charge on the polythene is given by $q = 4 \times 10^{-7} \ C$.
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 $e = 1.6 \times 10^{-19} \ C$.
To find the number of electrons $N$,we use the formula $N = \frac{q}{e}$.
Substituting the values: $N = \frac{4 \times 10^{-7}}{1.6 \times 10^{-19}}$.
$N = \frac{4}{1.6} \times 10^{-7 - (-19)} = 2.5 \times 10^{12}$.
Thus,the number of electrons transferred is $2.5 \times 10^{12}$.

(B) 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 $(e = 1.6 \times 10^{-19} \text{ C})$.
Given,$q = 3.52 \times 10^{-7} \text{ C}$.
To find the number of electrons transferred $(n)$,we rearrange the formula: $n = \frac{q}{e}$.
Substituting the values: $n = \frac{3.52 \times 10^{-7}}{1.6 \times 10^{-19}}$.
$n = 2.2 \times 10^{12}$.
Thus,the number of electrons transferred is $2.2 \times 10^{12}$.

(B) The correct option is $B$.
According to the quantization of charge,the total charge $Q$ on a body is given by the formula $Q = n \times e$,where $n$ is the number of electrons transferred and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
Given:
$n = 10^{19}$
$e = 1.6 \times 10^{-19} \ C$
Calculation:
$Q = 10^{19} \times 1.6 \times 10^{-19} \ C$
$Q = 1.6 \times 10^{(19-19)} \ C$
$Q = 1.6 \times 10^0 \ C$
$Q = 1.6 \ C$
Since electrons (which are negatively charged) are removed from the neutral plate,the plate develops a deficiency of electrons,resulting in a net positive charge.
Therefore,the charge on the plate becomes $+1.6 \ C$.

(B) The charge of a proton on a body is $q_p = n_p e = 10^{26} \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^7 \ C$.
The charge of an electron on a body is $q_e = -n_e e = -10^{24} \times 1.6 \times 10^{-19} \ C = -1.6 \times 10^5 \ C$.
The total charge $Q$ on the body is the sum of the charges of protons and electrons:
$Q = q_p + q_e = (1.6 \times 10^7) - (1.6 \times 10^5) \ C$.
$Q = (160 \times 10^5) - (1.6 \times 10^5) \ C$.
$Q = 158.4 \times 10^5 \ C = 1.584 \times 10^7 \ C$.
Rounding to two significant figures,we get $Q \approx 1.58 \times 10^7 \ C$.

(D) The charge on the plastic rod is $q = -8 \times 10^{-7} \ C$. Since the rod becomes negatively charged,it has gained electrons.
Using the quantization of charge formula,$q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the magnitude of the charge of an electron.
$n = \frac{|q|}{e} = \frac{8 \times 10^{-7}}{1.6 \times 10^{-19}}$
$n = 5 \times 10^{12}$
Since the plastic rod gained a negative charge,electrons must have been transferred from the wool to the plastic rod.

(C) stationary charge is a charge at rest. According to electrostatics,a charge at rest produces only an electric field in the space surrounding it.
It does not produce a magnetic field because a magnetic field is generated only by moving charges (currents).
Therefore,the correct option is $C$.

(B) The quantization of charge is given by the formula $q = n e$,where $q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$.
To find the number of electrons $n$,we rearrange the formula: $n = \frac{q}{e}$.
Given the charge $q = 1 \ nC = 1 \times 10^{-9} \ C$.
Substituting the values: $n = \frac{1 \times 10^{-9} \ C}{1.6 \times 10^{-19} \ C}$.
$n = \frac{1}{1.6} \times 10^{10} = 0.625 \times 10^{10} = 6.25 \times 10^{9}$.
Therefore,the number of electrons present on the body is $6.25 \times 10^{9}$.

(C) The charge on the body is given by $q = -3.2 \mu C = -3.2 \times 10^{-6} \ C$.
According to the quantization of charge,$q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(e = -1.6 \times 10^{-19} \ C)$.
Therefore,the number of excess electrons is given by $n = \frac{q}{e}$.
$n = \frac{-3.2 \times 10^{-6} \ C}{-1.6 \times 10^{-19} \ C}$.
$n = 2 \times 10^{13}$.

(B) When a positively charged glass rod is brought near an uncharged metal sphere,the free electrons in the metal are attracted toward the rod,creating a redistribution of charge (polarization).
However,the sphere remains electrically neutral as a whole because no charge is added to or removed from the sphere.
Since the sphere is mounted on an insulated stand,there is no path for charge to flow to or from the ground.
When the glass rod is removed,the redistributed charges return to their original positions,and the net charge on the metal sphere remains zero.

(B) Let the original charges on spheres $A$ and $B$ be $q_{1}$ and $q_{2}$ respectively.
Let the distance between the two spheres be $r$.
The initial force between them is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$.
Since both spheres are identical,when they are brought into contact,the total charge is redistributed equally between them.
Thus,the new charge on each sphere is $q^{\prime} = \frac{q_{1} + q_{2}}{2}$.
The new force of repulsion between spheres $A$ and $B$ is $F^{\prime} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{\prime} q^{\prime}}{r^{2}} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(\frac{q_{1} + q_{2}}{2})^{2}}{r^{2}}$.
According to the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $q_{1}$ and $q_{2}$ where $q_{1} \neq q_{2}$,we have $(\frac{q_{1} + q_{2}}{2})^{2} > q_{1} q_{2}$.
Therefore,$F^{\prime} > F$.

(B) Total number of atoms $N = 6 \times 10^{24}$.
Number of atoms from which one electron is removed is $n = 0.005 \% \text{ of } N$.
$n = \frac{0.005}{100} \times 6 \times 10^{24} = 5 \times 10^{-5} \times 6 \times 10^{24} = 30 \times 10^{19} = 3 \times 10^{20}$.
Since one electron is removed from each of these atoms,the total number of electrons removed is $3 \times 10^{20}$.
The charge gained by the solid is $q = n \times e$,where $e = 1.6 \times 10^{-19} \ C$.
$q = (3 \times 10^{20}) \times (1.6 \times 10^{-19}) = 4.8 \times 10 = 48 \ C$.
Since electrons are removed,the solid gains a positive charge of $+48 \ C$.

(C) According to the principle of quantization of charge,the total charge $Q$ on any body must be an integral multiple of the elementary charge $e$ (where $e = 1.6 \times 10^{-19} \ C$).
Mathematically,$Q = ne$,where $n$ is an integer $(n = 1, 2, 3, ...)$.
For option $A$: $n = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2$ (Integer).
For option $B$: $n = \frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 4$ (Integer).
For option $C$: $n = \frac{9.6 \times 10^{-20}}{1.6 \times 10^{-19}} = 0.6$ (Not an integer).
For option $D$: $n = \frac{9.6 \times 10^{-18}}{1.6 \times 10^{-19}} = 60$ (Integer).
Since $n$ must be an integer,a charge of $9.6 \times 10^{-20} \ C$ cannot exist.

(A) The charge on any body can be expressed as an integral multiple of the basic unit of charge,i.e.,the charge on one electron. This phenomenon is called the quantization of electric charge.
It can be written as $q = \pm ne$,where $n = 1, 2, 3, \dots$.
Charge is said to be quantized because it can have only discrete values rather than any arbitrary value. For example,a particle can have a charge of $+10e$ or $-6e$,but not a charge of $3.57e$.
From the above discussion,it is clear that electric charge is quantized,which means charge can only exist as an integral multiple of the charge of an electron. Therefore,half of the charge of an electron cannot exist.
Hence,both the assertion and the reason are correct,and the reason is the correct explanation of the assertion.

(A) Both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.
Electromagnetic forces are indeed much stronger than gravitational forces at the microscopic level. However,matter is composed of both positive and negative charges,which tend to cancel each other out,resulting in overall electrical neutrality for large objects.
Because of this cancellation,the net electromagnetic force between large,neutral objects is negligible. In contrast,gravitational force is always attractive and cumulative,meaning it does not cancel out. Therefore,on a cosmic scale,gravity becomes the dominant force governing the structure and formation of galaxies.

(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 $(e \approx 1.6 \times 10^{-19} \ C)$.
Given $Q = 2 \ C$.
Rearranging the formula for $n$: $n = \frac{Q}{e}$.
Substituting the values: $n = \frac{2}{1.6 \times 10^{-19}} \ C$.
$n = 1.25 \times 10^{19}$ electrons.