Electric Charge and Force

 Electrostatics

Introduction:$\\$The study of charges at rest is called static electricity or electrostatics.$\\$Charges are produced due to actual transfer of electrons.$\\$Positive charge means the deficiency of electrons while negative charge means excess (gain) of electrons.$\\$So, during charge transfer there is change in mass too. Thus, mass of -vely charged body is greater than that of -vely charged body.$\\$The attraction and repulsion of two charged objects are sometimes summarized as “Like charges repel, and opposite charges attract.” But keep in mind that the phrase “like charges” does not mean that the two charges are exactly identical, only that both charges have the same algebraic sign (both positive or both negative). “Opposite charges” means that both objects have an electric charge, and those charges have different signs (one positive and the other negative).

Properties of charges:

Quantization of charge$\\$Electric charge can exist only as an integral multiple of charge on an electron (-e) i.e.$\\$q= +ne, where ‘n' is an integer. The possible values of electric charges are$\\$q=±1e,±2e,±3e,…………$\\$Charge less than the charge on an electron (i.e. e=1.6×*$10^{-19}$ C ) is not possible.$\\$

Conservation of charge$\\$(a) On an isolated system, total electric charge always remains constant.$\\$(b) Total charge on a body is equal to algebraic of all the charges present on it. Every atom is electrically neutral as it contains as many electrons as the number of protons in it.$\\ \\$When we rub a glass rod with a piece of silk, the +ve charge acquired by the glass rod is equal to -ve charge acquired by silk piece. Thus charges are produced in equal and unlike pairs.$\\$Like charges repel each other while unlike charges attract each other. Repulsion is sure test of electrification. A charged body may attract a neutral body or an oppositely charged body but it always repels a similarly charged body.$\\$The magnitude of charge is not affected by its motion like mass i.e. charge is invariant. At very high speed (v=c), it is found that mass of a particle becomes m=$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \\$Where $m_0$ is the rest mass of particle.$\\$A charge at rest produces only electric field around itself, A charge having unaccelerated (uniform) motion produces$\\$electric as well as magnetic field around it. While a charge having accelerated motion emits electromagnetic radiation also in addition to producing electric and magnetic fields.

Mode of Charging:

Friction$\\$In this method body is charged by rubbing one surface with another and charge is induced .$\\$It is due to the transfer of electron from one body another by the thermal effect.$\\$

Induction$\\$

Conduction → By contact$\\$Charging a body by induction is preferable since the same charged body can be used to charge any$\\$number of bodies without loss of charge.$\\$

If q is the inducing charge, then charge induced on a body having dielectric constant $ε_r$ is given by $q_{induced} = q_{inducing} (1-\frac{1}{ε_r}$)$\\$where $ε_r$ is dielectric constant of uncharged body.$\\$-ve sign represents opposite nature of$\\$induction. For air or vacuum, $ε_r$ or K=1$\\$for conductor, $ε_r$ = ∞$\\$for insulator $ε_r$>1$\\$Case I: When uncharged body is conductor.$\\$K=∞$\\$$q_{induced} = q_{inducing} (1-\frac{1}{∞}) \\$Thus for conductor $q_{induced} = q_{inducing} \\$Case II: When uncharged body is insulator$\\$i.e. K>1$\\$•Thus, q induced < $q_{inducing}$$\\$Case III: For air$\\$K or $ε_r$=1$\\$$q_{induced}$=0$\\$So, there is no induction in air.$\\$###Conclusion:$\\$$q_{induced}≤q_{inducing}$$\\$In induction process, net induced charge on a body will be zero. In induction process,$\\$both charge and mass of charge body remain same but potential of charging body decreases.$\\$

Types of charge carriers:$\\$Substances are classified into 3 categories on the basis of flow of charge or electricity through them.$\\$Conductors: These easily allow electricity to pass through them.$\\$E.g.:-Metals, Earth, human body, etc.$\\$Insulators: These do not allow electricity to pass through them.$\\$E.g.:- Wood, Mica, glass, paper, etc.$\\$

Semiconductors: These lie in between conductors and insulators in their ability to conduct electricity.$\\$

E.g.:- Silicon and Germanium.$\\$

Charge Density:$\\$Linear charge density(𝛌)$\\$Charge per unit length on linear object called linear charge density and denoted by λ.$\\$Its unit is coulomb per meter (C$m^{-1}$ )$\\$$λ=\frac{q}{l} \\$**Surface charge density (σ) **$\\$Charge per unit area is known as surface charge density.$\\$Its unit is coulomb per$\\$square meter (C$m^{-2}$ ) i.e.$\\$$σ=\frac{q}{A} \\$Surface charge density on a charged conductor decreases with increase of radius of curvature and vice-versa.$\\$It is highest at the sharpest point of a conductor.$\\$Volume charge density ( ρ)$\\$Charge per unit volume of any charged bulk matter is called volume charge density.$\\$Its unit is coulomb per cubic meter (C$m^{-3}$ )$\\$$ρ=\frac{q}{V} \\$

Coulombs Law:

Description$\\$The force of attraction or repulsion between two point charges is directly proportional to the product of charge$\\$and inversely proportional to the square of the distance between them.$\\$■F∝$\frac{q_1.q_2}{r^2}$$\\$or,F=$\frac{q_1.q_2}{4πε_or^2} \\$Where $\frac{1}{4πε_0}=9×10^9 Nm^2 C^{-2}$ and$\\$$ε_o =8.85×10^{-12}{C^2}$ or$\\$Farad/meter is called permittivity of free space.$\\$■$F_{medium} =\frac{1}{4πε}\frac{q_1.q_2}{r^2} \\$So,$ε_r=\frac{F_{air}} {F_{medium}} " where " ε_r=\frac{ε}{ε_0}$ is relative Permittivity of the medium. $ε_r$ is also known as dielectric constant (K) of the medium.$\\$It has no units and no dimensions.$\\$K for metal is infinity.$\\$Dielectric constant is one for vacuum nearly one for air and zero for perfect insulator.$\\$$ε_r$ is unitless and dimensionless.$\\$

When two charges are held in air at distance r, then$\\$

■$F_{air} =\frac{1}{4πε_0}\frac{q_1.q_2}{r^2}=9×10^9 \frac{q_1.q_2}{r^2}) \\$[∵$ε_r$ for air =1]$\\$Coulomb’s force is valid only for point charges.$\\$The magnitude of Coulomb’s force at the location of two charges is always equal even charges are equal or different,$\\$but acts in opposite direction i.e.$F_1=-F_2$ (Newton’s $3^{rd}$law)$\\$■$\frac{F_1}{F_2}$ =1:1$\\$$F_1=F_2=\frac{1}{4πε_0}\frac{q_1.q_2}{r^2} \\$Coulomb’s force doesn’t depend on the mass of charge i.e. the magnitude of Coulomb’s force at the location of two charges is always same even the mass of the charges are different$\\$a=$\frac{F}{m} \\$or, a∝$\frac{1}{m}$ [ F is constant ]$\\$or, $\frac{v}{t}$∝$\frac{1}{m} \\$∴v∝$\frac{1}{m} \\$⇒$\frac{v_2}{v_1}$=$\frac{m_1}{m_2 } \\$Thus, the speed gained by a charge only depends on the mass of the charge.$\\$

Relation between different forces:$\\$$F_{Coulomb}$ =$\frac{q_1.q_2}{4πε_or^2} \\$(1)Proton - Proton …$\\$$F_{gravitational} =\frac{G.m_1.m_2}{r^2}$$\\$$\frac{F_G}{F_C} =10^{-36} \\$(2)Proton – Electron$\\$$\frac{F_G}{F_C} =\frac{4πε_0 G.m_1.m_2}{q_1.q_2}$$\\$$\frac{F_G}{F_C} =10^{-30}$$\\$∴$F_C>>>F_G$$\\$(3)Electron - Electron$\\$$\frac{F_G}{F_C} =10^{-42} \\$If $F_G=F_C \\$1=$\frac{4πε_0 Gm^2}{q^2}$ (If particles are identical)$\\$$\frac{q}{m}=\sqrt{4πε_0G} \\$$F_nuclear >F_C>F_G \\$Inside the nucleus, both nuclear and electrostatic force exist.$\\$

Note: Gravitational force is neglected for proton, electron and neutron in comparison to Coulomb and Nuclear forces.

Van de graff generator is a powerful machine used for generating high positive potentials ≈$10^6$Volt.

Dielectrics are of 2 types: Non Polar and Polar.$\\$(1) The non polar dielectrics: (like$N_2,O_2$, Benzene, Methane) etc. are made up of non-polar atoms/molecules,$\\$in which centre of mass of positive charge coincides with the centre of mass of negative charge of the atom/molecule.$\\$(2) The polar dielectrics (like $H_2O,CO_2,NH_3$, HCl) etc. are made up of polar atoms/molecules, in$\\$which the centre of mass of +ve charge does not coincide with the centre of mass of -ve charge of the atom/molecule.$\\$

• A non polar dielectric can be polarized by applying an external electric field on the dielectric.

POINTS

Two identical pith balls each of mass ‘m’ are charged with a charge ‘q’ each. If the two balls are$\\$suspended by a silk thread of length ‘l’ from the same hook as shown in the figure then the distance between$\\$the balls in equilibrium condition is given by$\\$$\frac{F}{mg} = tan θ = \frac{x}{2l} \\$

In the above problem, if the whole set up is taken in gravity free space, then the angle between the two strings is $180^∘$$\\$and the tension in each string is equal to $P=\frac{1}{4πε_0}\frac{q^2}{4l^2} \\$-> In the above case, if the balls are suspended in a liquid of density ρ and the distance between the balls remains same, then the dielectric constant of the liquid is given by$\\$K=$ε_r$=$\frac{ρ'}{ρ'-ρ} \\$Also $\frac{F}{mg} = tanθ = \frac{F_{medium}}{mg(1-\frac{ρ}{ρ'})}$$\\$

			Where ρ' = density of the material of the ball.$\\$

• Coulomb’s force may be attractive or repulsive but gravitational force is always attractive.$\\$

• Coulomb’s electrostatic force depends on the medium between the two charges but gravitational force is independent of the medium between the two bodies.$\\$

$F_{air}=\frac{q_1.q_2}{4πε_o.r^2}$ and$\\$

$\frac{F_{air}}{F_{med}} =\frac{ε_0⋅ε_r}{ε_0} =ε_r$ or K.$\\$$\frac{F_{air}}{F_{med}}$=K:1$\\$K for conductor is infinity and for insulator it is > 1.$\\$∴$F_{air} >F_{medium} \\$Thus, coulomb’s force will be maximum in air or vacuum medium and with any other insulator; Coulomb’s force will be less than that of air.$\\$If two charges are placed at separation 'r and dielectric constant ‘K’ is placed in between, then$\\$

F=$\frac{q_1 q_2}{4πε_0[r-t+t\sqrt{K}]^2} \\$Similarly, when ‘n’ numbers of slabs are placed, then$\\$F=$\frac{q_1.q_2}{4πε_0 [t_1 \sqrt{K_1}+t_2 \sqrt{K_2}+t_3\sqrt{K_3}+….....]^2 } \\$

Two charges are placed in vacuum and in a medium of dielectric constant K. To have the same force between the two charges in vacuum and in the medium,$\\$their separation is given by $r_m=\frac{r_0}{\sqrt{K}} \\$If a charge Q is divided into two parts such that the force between them is to be maximum, then each part has charge equal to Q/2$\\$When two identical conductors having charges ‘$q_1$' and ‘$q_2$ ’ are put in contact and then separated, then each conductor has a charge equal to $\frac{(q_1+q_2)}{2}$.$\\$

• Similarly in the above statement, if the charges on these bodies are $q_1$ and

-$q_2$ then each has a charge $\frac{(q_1-q_2)}{2}$ after sharing$\\$If 3 identical charges (+q) are placed at 3 corners of equilateral triangle, then resultant force on any one charge due to rest two charges is F=√3 $\frac{q^2}{r^2}$$\\$i. Now if a fourth charge Q=-q is placed at centroid all three charges move to centroid.$\\$ii. But if the charge at centroid Q=q/3, all four charges will remain stationary.$\\$

For a square with four identical charges +q on four corners, the charge at centre should be$\\$Q =-$\frac{q}{4(1+2√2)}$ for the system to be in equilibrium (i.e. $F_{net}$ = 0).$\\$[Hint: Four charges at the comer of square experiences equal force F=$\frac{q^2}{4πε_0 r^2} \\$Resultant of force due charges at C and D and charges at B and D is F' which lies in same direction of force due to charges at A and D and charges at O and D.$\\$F'=2Fcos$\frac{θ}{2} \\$=2Fcos⁡$45^∘$=√2 F$\\$$F_1=\frac{q^2}{4πε_0(√2l)^2}$ [∵ AD=√2l ]$\\$$F_2 = \frac{q^2}{4πε_0(l/√2)^2} \\$$F_{net}=F_1 + F_2 \\$=$\frac{q}{4πε_0l^2} [2Q$+$\frac{q}{2}+√2 q]$=0$\\$∴Q=-q($\frac{1+2√2}{4}$)$\\$

Similarly for triangle resultant of force due to charges at A and C and charges at B and C is F' which lies in same line with force due to charges at O and C($F_3$)$\\$$F_1=F_2=\frac{q^2}{4πε_0l^2} \\$F=$\sqrt{F^2+F^2+2F^2cos60^o} \\$=√3 F$\\$$F_3=\frac{Qq}{4πε_0l^2} \\$∴$F_{net}=F'+√(3F) = 0 \\$∴Q = -q/√3$\\$A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if 'q' is equal to'-Q/4'.$\\$[Hint: For system to be in equilibrium net force on every charge should be zero.]$\\$Considering charge at A$\\$$\frac{1}{4πε_0}\frac{Qq}{x^2}$ +$\frac{1}{4πε_0}\frac{Q⋅Q}{(2x)^2}$=0$\\$∴q=-$\frac{Q}{4}$$\\$

For this type of problems, if charge at the ends is +ve then charge at the centre is –ve and if charges at A and B are -ve charge at centre is +ve and should be divided by 4. For eg:$\\$

Applying the trick$\\$x=$\frac{e}{4}$ [Since charges at ends are -ve].$\\$

• For like charges, a third charge experience zero force when kept between the charges nearby lower charge.$\\$
• For unlike charges, a third charge experiences zero force outside the charges nearby lower charge.$\\$

Note: While checking higher or lower charge check only magnitude.$\\$

• Distance of third charge from higher charge, where net force experience by it is zero.$\\$

x=$\frac{distance between charges}{1±\sqrt{\frac{lower charge} {higher charge}}} \\$

• for like charges and -for unlike charges.$\\$

The ratio of forces between two small spheres with constant charge in air and in a medium of dielectric constant K is K:1.$\\$The ratio of the forces between two small spheres charged to constant potentials in air and in medium of dielectric constant K is K:1.$\\$Divergence of gold leaf electroscope (GLE) can be used to study charge.$\\$

• When the body to be tested is brought to the disc of the GLE, if the leaves of GLE do not diverge, the body must be uncharged and if the leaves of GLE show divergence, the body must be charged.$\\$
• If a soap bubble is given small +ve or -ve charge, its radius increases because of repulsion and its surface tension decreases.$\\$
• The phenomenon of acquiring temporary electrification under the influence of a charged body is known as electric induction.$\\$
• Faraday’s icepail experiment establishes relationship between inducing and induced charge.$\\$

Result:$\\$

i. Induced + ve charge = Induced -ve charge$\\$ii. Induced charge = Inducing charge