Coulomb's Law Electro Statics

Mohammed Asif

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Topic : Electrostatics

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Coulomb’s Law

ELECTROSTATICSThe branch of physics dealing with charges at restUnder electrostatic condition, the electric field inside the metallic conductor remains zero.Application in instruments like microphones, cathode ray tube, capacitors, van de Graaff generator, photocopier, electrostatic shielder etc.

CHARGEBenzamin franklin and MilikanFundamental property exhibited due to deficiency or sufficiency of electronsTwo types of charges : Positive and negative chargeUnlike charge attract one another and like charges repelExhibit properties like Quantization, superposition, invariance, conservation, induction etcNegatively charged bodies : electron sufficient & slightly mass sufficient also

QUANTISATIONQ = wher n is intergere = - 1.6 x 10-19CException : Quark particle Not continuous transfer but in discrete packet

OBJECTIVE 1How many electrons are there in 1C of charge?i) 1 ii) 6.25 x 1018 iii) 1.6 x 10-19 iv) None

Solution: (ii)

OBJECTIVE 2Minimum charge that a body can retain isi) 1.6 x 10-19C ii) 3.2 x 10-19 iii) 2.4 x 10-19C iv) None

Solution: (i)The minimum charge that a body can retain is one electronic charge

SUBJECTIVE 3

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What would be the interaction force between two copper spheres, each of mass 1gm, separated by the distance 1m, if the total electronic charge in them differs from the total charge of the nuclei by one percent?

Solution:Atomic weight of

No. of atoms in 1gm

No. of electrons

Charge

(Since 1% of electrons has been transferred)= 4.39 x 102C

Electric force between two charged particle Inversely proportional to the square of separation between the two particles and directed

along the line joining charges Proportional to the product of the magnitudes of charges q1 & q2 on the two particles Attractive if charges are of opposite sign and repulsive if having same sign

Mathematically,

Where is known as permittivity of free space.

Victorial form

Force on 2 due to

should be in direction of Colombian force.

Force on 2 due to



OBJECTIVE 4Calculate the ratio of electrostatic to gravitational interaction between two electrons

Solution:

Electrostatic force is much stronger compared to gravitational force.

OBJECTIVE 5A point charge q is situated at a distance ‘d’ from one end of a thin non conducting rod of length L having a charge Q (uniformly distributed along its length) as shown in figure. Find the magnitude of electric force between the two.

Solution:Consider an element of rod, of length d x, at a distance x from the point charge q as shown in figure. Treating the element as a point charge, the force between q and charge element will be as:

Charge on elementary strip

Coulombian force acting on strip

Therefore,



SUBJECTIVE 6A thin wire ring of radius R has an electric charge q. What will be the charge of the force stretching the wire if a point charge q0 is placed at the ring’s centre?

Solution:When the wire is given charge q, it comes in tension, called hoop tension. When another charge q0 is placed at the centre of the ring, the tension in the wire increases to keep this segment into equilibrium again.

Consider an elementary strip subtending angle d at its centre.

The string is in equilibrium in X-direction. For equilibrium in y-direction,

……………(i)

Charge on elementary strip

…………….(ii)

Substituing (ii) in (i)

SUBJECTIVE 7The identical beads each having a mass m and charge q, when placed in hemi spherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are at a distance R apart. Determine the charge on each bead.



Solution:The bead is under electrostatic equilibrium and force acting are its weight, electrostatic force and Normal Reaction.

For equilibrium N sin 600 = mg ………..(i)

………..(ii)

Dividing (i) by (ii),

SUPERPOSITIONThe interaction between any two charges completely unaffected by the presence of other chargesNet resultant force on any charge equals vectorial sum o coulombian force contributed by individual particle in surroundings

Net force on q1 =

All vectorial technique to be applied for calculation of net force.For three dimensional distribution of charges, the coulomb’s law in vectorial form should be used.

OBJECTIVE 8Three charges q are placed at the vertices of equilateral triangle of side a. Find the electrostatic force on the charge Q kept at the centroid of triangle.



Solution :

Since three equal forces are acting at a point making angle of 1200, net resultant force is zero. Concept : The net force acting on the charge kept at centre of regular polygon is zero due to charges kept at the vertices provided they are of same magnitude & sign.

OBJECTIVE 9Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of side Lm. The magnitude of the force on the point charge of –q coulomb placed at the centre of hexagon will be:i) Zero ii) iii) iv) None of these

Solution :If there would have been charge at the left vertices also, net force at the centre will be zero.

The coulombian force due to five charge should balance the coulombian force exerted by the charge dept at left vertices. Force due to all charges should be along OA.

OBJECTIVE 10There equal charges, each equal to q, are placed at the vertices of an equilateral triangle of side a find the force on the left most corner of charge q.

Solution:



The net resultant force= F1

OBJECTIVE 11A charge Q is placed at each of the opposite corner of a square. A charge q is placed at each at each of the two other two corners. If the resultant force on Q is zero, then:a) b) c) d)

Solution:

Net resultant =

OBJECTIVE 12A cube of edge ‘a’ carries a point charge q at each corner. Show that the magnitude of the resultant force on any one of the charge is

Solution:Refer to coulomb’s law in vector form



According to superposition principle,

SUBJECTIVE 13

Two point charges A and B have charges respectively and position vectors and

respectively. Find the force on the charge at A due to B.Solution :

Force on A and to B



STRING RELATED PROBLEMSDraw the free body diagram of forceWrite down the condition for equilibriumSolve the equation after co-relating situation.

SUBJECTIVE 14Two small charged spheres, each of charge q and each of mass m, are suspended form the same point by silk threads of length 1(a) Find the distance between the spheres x, if x < < < 1, for the equilibrium of charges.

(b) If the charge starts leaking form both the spheres at the constant rate , find the velocity

of approach of the spheres.Solution:

The force acting on the charged spheres are shown in the equilibrium.

………..(i)………..(ii)

Dividing (i) by (ii),

If x < < < 1, then

Or

ii)

Differentiating both side with respect to t



SUBJECTIVE 15If the two balls in the above mentioned problem contains different charges q1 > q2. The deflection in the cases are as shown in figure. The two balls are of same mass, then

i) ii) iii) iv) Cannot be predicatedSolution: (iii)

since two balls are of same mass and electrostatic force is also same.

SUBJECTIVE 16Three small balls, each of mass m are suspended separately from a common point by silk threads, each of length 1. The balls are identically charged and hung at the corners of equilateral triangle of side a (a < < < 1). What is the charge on each ball?

Solution:The lengths of the threads are same, the mass of each ball is same, the balls are identically charged, and the balls are suspended form the same point. Under the equilibrium, the balls will make an equilateral triangle in a horizontal Plane.Let ABC be the triangle formed by them.

The force on one of the balls at c are (i) Coulombic repulsion by charges at A & B

ii) mg, acting down wardiii) Tension in thread

…………(1)…………(2)



From (1) and (2)

…………(3)

If tan is small, then

…………(4)

Substituting (4) in equal no. (3),

SUBJECTIVE 17A particle A having a charge of 5 x 10-7C is fixed in a vertical wall. A second particle B of mass 100g and having equal charge is suspended by a silk thread of length 30cm form the wall. The point of suspension is 30cm above particle A. Find the angle of the thread with the vertical when it stays in equilibrium?

Solution:

So the electrostatic force between the charge will

Be …………….(i)

Lami’s theorem can be applied in order to solve this question



Or,

…………….(ii)

Equating (i) and (ii),

In the given problem,q = 5 x 10-7C

Or

CONSERVATIONTotal charge of isolated system remains constant,Appearance of charge is due to slight disturbance of electrical neutrality of each.

OBJECTIVE 18A certain charge Q is to be divided into two parts, q and (Q-q). What is the relationship of Q to q of the two parts, placed at a given distance apart are to have maximum coulomb repulsion?

Solution:

For F to be maximum, ,

SUBJECTIVE 19Two identical oppositely charged metallic sphere placed 0.5m apart attract each other with a force of 0.108 N. When they are connected by a Cu wire, they begin to repel each other with a force of 0.036 N. What are the initial charges on them?

Solution:Let the charge be q1 and q2 before connecting by cu-wire. After connection, charges on each will

be

………..(i)



…………(ii)

After solving these two equations,

OSCILLATIONSUnder equilibrium condition, net force on particle = 0By displacing slightly, the net force should be restoring in nature.If net force displacement, it will be making SHM

Exampls

SUBJECITIVE 20Two particles A and B, each carrying a charge Q, are held fixed with O separation 2l between them. A particle C having mass m and charge q is kept at the middle points of the line AB. If the particle c is displaced through a distance x and x <<<1, then find the condition of SHM, oscillation and also the time period of oscillation. Condition –I: If it is displaced along the line AB Condition –II: If it is displaced perpendicular to the line AB.

Solution:Condition –I: The particle has to be positive in sign for oscillation along the axis.

Suppose the particle C is displaced by small distance x.

Net force towards the centre = FB - FA

If x <<< 1, then oscillation will become SHM.



Condition-II: The particle C has to be negative in sign in order to have restoring force along perpendicular bisector.

Net restoring force =

If x <<< l, oscillation will become SHM

or

Or

Equatorial oscillation is having time period times compared to axial oscillation.

SUBJECTIVE 21Three particles each carrying charge are placed at the vertices of triangle of side ‘a’ A fourth particle of charge –q and mass m is kept at the centroid of the triangle and then displaced by very small distance x along the vertical direction.Find the time period of oscillation.

Solution :When charged particle is kept at centroid, the net electrostatic force on it will be zero. Taking centroid as origin of Co-ordinate system, the co-ordinates of points is as follows:

Taking vector from of coulomb’s law



If x <<< a,

;

SUBJCTIVE 22Two free particles with charge +q and +4q are at a distance L apart. A third charge is placed so that the entire system is in equilibrium. Find the location, magnitude and sign of third charge?

Solution: Third charge can be of any sign for its equilibrium. But it has to be negative in order to make force on q and 4q equal to zero. Then only the entire system will be in equilibrium. For net force on –Q1 to be zero.

For net force on q to be zer,

EFFECT OF MEDIUMIn medium the electrostatic force reduces by factor of K (dielectric constant)

If charge is kept in interfacing media, then effective distance has to be calculated for calculation of force

Effective distance



Force in interfacing media

OBJECTIVE 23If charge q1 and q2 kept in interfacing media of three different dielectric constant 9, 81 and air. The thickness of each median in interfacing system is equal. If the distance between the charge is r. Then the coulomb force acting between than.

Solution:The effective distance in medium Net effective distance

Colombian force

SUBJECTIVE 24Two balls each having a density d are suspended form Q common point by two insulated string of equal length. Both the balls have same mass and charge. In equilibrium, each string makes angle with the vertical. Both the balls are immersed in a liquid. As a result, angle does not change. The density of liquid is . Find the dielectric strength?

Solution:Refer to concept section,

…….. (i)

When the system is immersed in liquid medium, then the force (electrostatic) becomes F/K and upward thrust acts on the system.

…….. (ii)



Equating (i) and (ii)



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Coulomb's Law Electro Statics