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Friday, 21 April 2017

Electrostatics, Chapter Notes, Class 12, Physics, (IIT-JEE & AIPMT)

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 (a) Introduction : Electromagnetism is a science of the combinatin of electrical and magnetic phenomenon. Electromagnetism can be divided into 2 parts : (1) Electrostatics : It deals with the study of charges at rest. (2) Electrodynamics : It deals with the study of charges in motion (discusses magnetic phenomenon). In this chapter we will be dealing with charges at rest i.e. electrostatics.   (b) Structure of Atom : An atom consists of two parts (i) nucleus (ii) extra nuclear part. Nucleus consists of neutrons and protons and extra nuclear part has electrons revolving around nucleus. In a neutral atom. number of electrons = number of protons. charge of electrons = charge of protons = 1.602 × 10-19 coulomb. Normally positive charges are positron, proton and positive ions. In nature practically free existing positive charge are positive ions and negative charges are electrons.   (c) Electric Charge Charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some of naturally charged particles are electron, proton, a-particle etc.   (d) Types of Charge (i) Positive charge : It is the deficiency of electrons compared to protons. (ii) Negative charge : It is the excess of electrons compared to protons.   (e) Units of Charge Charge is a derived physical quantity. Charge is measured in coulomb in S.I. unit. In practice we use mC  (10-3C), mC (10 -6 C), nC (10-9C) etc. C.G.S unit of charge = electrostatic unit = esu. 1 coulomb = 3 × 109 esu of charge Dimensional formula of charge = [M°L°T1I1]   (f) Properties of Charge   (i) Charge is a scalar quantity : It adds algebrically and represents excess, or deficiency of electrons. (ii) Charge is transferable : Charging a body implies transfer of charge (electrons) from one body to another. Positively charged body means loss of electrons, i.e. deficiency of electrons. Negatively charged body means excess of electrons. This also shows that mass of a negatively charged body > mass of a positively charged identical body. (iii) Charge is conserved : In an isolated system, total charge (sum of positive and negative) remains constant whatever change takes place in that system. (iv) Charge is quantized : Charge on any body always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of charge on electron (1e = 1.6 × 10-19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan's oil drop experiment proved the quantization of charge or atomicity of charge. Recently, the existence of particles of charge ± and ± has been postulated. These particles are called quarks but still this is not considered as the quantum of charge because these are unstable (They have very short span of life.) (v) Like point charges repel each other while unlike point charges attract each other. (vi) Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. The particle such as photon or neutrino which have no (rest) mass can never have a charge. (vii) Charge is relativistically invariant : This means that charge is independent of frame of reference, i.e., charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charge, the mass of a body depends on its speed and increases with increase in speed. (viii) A charge at rest produces only electric field around itself; a charge having uniform motion produces electric as well as magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation.   (g) Conductors and Insultators : Any object can be broadly classified in either of the following two categories : (i) Conductors (ii) Insulators (i) Conductors : These are the materials that allow flow of charge through them. This category generally comprises of metals but may sometimes contain non-metals too. (ex. Carbon in form of graphite.) (ii) Insulators : These are the materials which do not allow movement of charge through them.   (h) Charging of Bodies : An object can be charged by addition or removed of electrons from it. In general an object can either be a conductor or insulator. Thus we are going to discuss the charging of a conductor and charging of an insultor in brife. (i) Charging of Conductors : Conductors can be charged by (a) Rubbing or frictional electricity (b) Conduction & Induction (will be studied in later sections) (c) Thermionic emission (will be study the topic "heat") (d) Photo electric emission (will be studied under the topic modern physics) (ii) Charging of Insulators : Since charge cannot flow through insulators, neither conduction nor induction can be used to charge, insultators, so in order to charge an insulator friction is used. Whenever an insulator is rubbed against a body exchange of electrons takes place between the two. This results in apperance of equal and opposite charges on the insulator and the other body. Thus the insulator is charged. For example rubbing of plastic with fur, silk with glass causes charging of these things.   2. Coulomb's Law : Coulomb, through his experiments found out that the two charges 'q1' and 'q2' kept at distance 'r' in a medium as shown in figure-1 exert a force 'F' on each other. The value of force F is given by This law gives the net force experienced by q1 and q2 taking in account the medium surrounding them. Where, F gives the magnitude of electrostatic force. q1 and q2 are the magnitudes of the two interacting charges. K is electrostatic constant which depends upon the medium surrounding the two charges. This force F acts along the line joining the two charges and is repulsive if q1 and q2are of same sign and is attractive if they are of opposite sign. Let us take some examples on application of coulomb's Law.       2.1 Vector forms of Coloumb's law  (F21 : force on q2 due to q1)    Head of points at that position where force has to be calculated.   &  depend on origin but  does not.  q1 and q2 should be put along with sign.   2.2 Coloumb's law in a medium :   (i) Relative Permittivity When two charges are placed in vacuum or when the same set of charges are placed in a medium, the net force experienced by the charges will be different. The effect of presence of medium is accounted in the proportionality constant K. This electrostatic constant K is defined as K =  where Î = Î0 Îr where Î = absolute permittivity of medium Î0 = permittivity of free space. having a constant value = 8.85 × 10-12 coul2/N-m2 = relative permittivity of medium with respect to free space, also termed as dielectric constant. For free space Îr = 1 and  = 9 × 109  (ii) Force dependency on Medium We can say that when two charges are placed in vacuum (or air) the force experienced by the charges can be given as When these charges are submerged in a medium, having dielectric constant Îr, the force becomes Fmed =  or Fmed =  as Îr > 1 ⇒ Fmed < Fair          3. Electric field : The figure shown a charge q is lying in free space. Now a charge q¢ is brought near it. By columb's law we know that the charge q experiences a force and it exerts an equal force on q¢. How does q become aware of the presence of q¢ ??? (We don't expect q to have sensory organs just as we have) The answer is electric field !!! Electric field is the space surrounding an electric charge q in which another charge q¢ experiences a (electrostatic) force of attraction, or repulsion.   The direction of electric field is radially outwards for a positive charge and is radially inwards for a negative charge as shown in the figure above. There are some points always to be kept in mind. These are (1) Electric field can be defined as a space surrounding a charge in which another static charge experiences a force on it. (2) In a region electric field is said to exist if an electric force is exerted on a static charge placed at that point. (3) It is important to note that with every charge particle, there is an electric field associated which extends up to infinity. (4) No charged particle experiences force due to its own electric field.  =   A very small positive charge which does not produce its significant electric field is called a test charge. Thus electric field strength at point can be defined generally as "Electric field strength at only point in space to be the electrostatic force per unit charge on a test charge." If a charge q0 placed at a point in electric field, experiences a net force on it, then electric field strength at that point can be or  ....(1) [q0 ® test charge]     (a) Electric Field Strength due to Point Charge : As discussed earlier, if we find electric field due to a point charge at a distance x from it. Its magnitude can be given as  ...(2) (b) Vector Form of Electric field due to a Point Charge : As shown in figure, the direction of electric field strength at point P is along the direction of . Thus the value of  can be written as or  ...(3)  It should be noted that the expression in equation (2) and (3) are only valid for point charges. We can not find electric field strength due to charged extended bodies by concentrating their whole charge at geometric centre and using the result of a point charge.   3.1 Graph of electric field due to binary charge configuration 1.  2.  3.  4.    3.2 Electric field Strength at a General Point due to a Uniformly Charged Rod : As shown in figure, if P is any general point in the surrounding of rod, to find electric field strength at P, again we consider an element on rod of length dx at a distance x from point O as shown in figure. Now if dE be the electric field at P due to the element, then it can be given as Here dq =  Now we resolve electric field in components. Electric field strength in x-direction due to dq at P is dEx = dE sin q or  =  Here we have x = r tan q and dx = r sec2 qdq Thus we have  Strength =  Net electric field strength due to dq at point P in x-direction is or  or  Similarly, electric field strength at point P due to dq in y-direction is dEy = dE cos q or dEy =  Again we have x = r tan q and dx = r sec2 q dq Thus we have dEy =  =  Net electric field strength at P due to dq in y-direction is Ey =  or Ey =  or Ey =  Thus electric field at a general point in the surrounding of a uniformly charged rod which subtend angles q1 and q2 at the two corners of rod can be given as in ||-direction Ex =  =  in ^ -direction Ey =  =   r is the perpendicular distance of the point from the wire  q1 and q2 should be taken in opposite sense   3.3 Electric field due to infinite wire (l >> r) Here we have to find the electric field at point p due to the given infinite wire. Using the formula learnt in above section which For above case,   Enet at    3.4 Electric field due to semi infinite wire For this case Enet at s   3.5 Electric field due to Uniformly Charged Ring : Case - I : At its Centre Here by symmetry we can say that electric field strength at centre due to every small segment on ring is cancelled by the electric field at centre due to the segment exactly opposite to it. As shown in figure. The electric field strength at centre due to segment AB is cancelled by that due to segment CD. This net electric field strength at the centre of a uniformly charged ring is zero   Case II : At a Point on the Axis of Ring For this look at the figure. There we'll find the electric field strength at point P due to the ring which is situated at a distance x from the ring centre. For this we consider a small section of length dl on ring as shown. The charge on this elemental section is dq =  dl [Q = total charge of ring] Due to the element dq, electric field strength dE at point P can be given as The component of this field strength dE sin a which is normal to the axis of ring will be cancelled out due to the ring section opposite to dl. The component of electric field strength along the the axis of ring dE cosa due to all the sections will be added up. Hence total electric field strength at point P due to the ring is =  or  =  =  EP =    3.6 Electric field Strength due to a Uniformly Surface Charged Disc : If there is a disc of radius R, charged on its surface with surface charge density s coul/m2, we wish to find electric field strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure. To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc as shown in figure. Now the charge on this elemental ring dq can be given as dq = s 2p y dy [Area of elemental ring ds = 2py dy] Now we know that electric field strength due to a ring of radius R. Charge Q at a distance x from its centre on its axis can be given as  [As done earlier] Here due to the elemental ring electric field strength dE at point P can be given as  =  Net electric field at point P due to this disc is given by integrating above expression from O to R as = Kspx  = 2Kspx  E =  Case : (i) If x >> R =  =  =  i.e. behaviour of the disc is like a point charge.   Case (ii) : If x << R i.e. behaviour of the disc is like infinite sheet.   3.7 Electric Field Strength due to a Uniformly charged Hollow Hemispherical Cup : Figure shows a hollow hemisphere, uniformly charged with surface charge density s coul/m2. To find electric field strength at its centre C, we consider an elemental ring on its surface of angular width dq at an angle q from its axis as shown. The surface area of this ring will be ds = 2pR sin q × Rdq Charge on this elemental ring is dq = sds = s. 2pR2 sin q dq Now due to this ring electric field strength at centre C can be given as =  = pKs sin 2q dq Net electric field at centre can be obtained by integrating this expression between limits 0 to as E0 =  =  =    4. Conservative Force A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and not on the nature of path followed between the initial and final positions.    Consider a body of mass m being raised to a height h vertically upwards as shown in above figure. The work done is mgh. Suppose we take the body along the path as in (b). The work done during horizontal motion is zero. Adding up the works done in the two vertical path of the paths, we get the result mgh once again. Any arbitrary path like the one shown in (c) can be broken into elementary horizontal and vertical portions. Work done along the horizontal path is zero. The work done along the vertical parts add up to mgh. Thus we conclude that the work done in raising a body against gravity is independent of the path taken. It only depends upon the intial and final positions of the body. We conclude from this discussion that the force of gravity is a conservative force.   Examples of Conservative forces. (i) Gravitational force, not only due to Earth due in its general form as given by the universal law of gravitation, is a conservative force. (ii) Elastic force in a stretched or compressed spring is a conservative force. (iii) Electrostatic force between two electric charges is a conservative force. (iv) Magnetic force between two magnetic poles is a conservative forces. Forces acting along the line joining the centres of two bodies are called central forces. Gravitational force and Electrosatic forces are two important examples of central forces. Central forces are conservative forces.   Properties of Conservative forces Work done by or against a conservative force depends only on the initial and final positions of the body. Work done by or against a conservative force does not depend upon the nature of the path between initial and final positions of the body. If the work done by a force in moving a body from an initial location to a final location is independent of the path taken between the two points, then the force is conservative. Work done by or against a conservative force in a round trip is zero. If a body moves under the action of a force that does no total work during any round trip, then the force is conservative; otherwise it is non-conservative. The concept of potential energy exists only in the case of conservative forces. The work done by a conservative force is completely recoverable. Complete recoverability is an important aspect of the work of a conservative force.   Work done by conservative forces Ist format : (When constant force is given) Ex.33 Calculate the work done to displace the particle from (1, 2) to (4, 5). if Sol. dw = ( ) dw =  ⇒ dw = 4dx 3dy  =     ⇒ w =  w = (16 - 4) (15 - 6) ⇒ w = 12 9 = 21 Joule II format : (When F is given as a function of x, y, z) If  then dw =  ⇒ dw = Fxdx Fydy FZdz   Ex.34 An object is displaced from position vector  to  under a force . Find the work done by this force. Sol.  =  Ans.   IIIrd format (perfect differential format)   Ex.35 If  then find out the work done in moving the particle from position (2, 3) to (5, 6) Sol. dw =  dw =  dw = ydx xdy Now ydx xdy = d(xy) (perfect differential equation) ⇒ dw = d(xy) for total work done we integrate both side Put xy = k then at (2, 3) ki = 2 × 3 = 6 at (5, 6) kf = 5 × 6 = 30 then w =  ⇒ w = (30 - 6) = 24 Joule   4.1 Non-conservative forces : A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path between the initial and final positions. The frictional forces are non-conservative forces. This is because the work done against friction depends on the length of the path along which a body is moved. It does not depend only on the initial and final positions. Note that the work done by fricitional force in a round trip is not zero. The velocity-dependent forces such as air resistance, viscous force, magnetic force etc., are non conservative forces.   Difference between conservative and Non-conservative forces   5. ELECTROSTATIC POTENTIAL ENERGY :   (a) Electrostatic Potential Energy : Potential energy of a system of particles is defined only in conservative fields. As electric field is also conservative, we define potential energy in it. Before proceeding further, we should keep in mind the following points, which are useful in understanding potential energy in electric fields. (i) Doing work implies supply of energy (ii) Energy can neither be transferred nor be transformed into any other form without doing work (iii) Kinetic energy implies utilization of energy where as potential energy implies storage of energy (iv) Whenever work is done on a system of bodies, the supplied energy to the system is either used in form of KE of its particles or it will be stored in the system in some form, increases the potential energy of system. (v) When all particles of a system are separated far apart by infinite distance there will be no interaction between them. This state we take as reference of zero potential energy. Now potential energy of a system of particles we define as the work done in assembling the system in a given configuration against the interaction forces of particles.   Electrostatic potential energy is defined in two ways. (i) Interaction energy of charged particles of a system. (ii) Self energy of a charged object (will be discussed later)   (b) Electrostatic Interaction Energy : Electrostatic interaction energy of a system of charged particles is defined as the external work required to assemble the particles from infinity to a given configuration.When some charged particles are at infinite separation, their potential energy is taken zero as no interaction is there between them. When these charges are brought close to a given configuration, external work is required if the force between these particles is repulsive and energy is supplied to the system hence final potential energy of system will be positive. If the force between the particles is attractive work will be done by the system and final potential energy of system will be negative. Let us take some illustrations to understand this concept in detail. (c) Interaction Energy of a System of Two Charged Particles :   5.1 Motion of a Charge Particle and Angular Momentum Conservation : We know that a system of particles when no external torque acts, the total angular momentum of system remains conserved. Consider following examples which explains the concept for moving charged particles. Ex.40 Figure shows a charge Q fixed at a position in space. From a large distance another charge particle of charge q and mass m is thrown toward Q with an impact parameter d as shown with speed v. find the distance of closest approach of the two particles. Sol.   Here we can see that as q moves toward Q, a repulsive force acts on -q radially outward Q. Here as the line of action of force passes through the fix charge, no torque act on q relative to the fix point charge Q, thus here we can say that with respect to Q, the angular momentum of q must remain constant. Here we can say that q will be closest to Q when it is moving perpendicularly to the line joining the two charges as shown. If the closest separation in the two charges is rmin, from conservation of angular momentum we can write mvd = mv0 rmin ...(1) Now from energy conservation, we have Here we use from equation (1) v0 =   (2) Solving equation (2) we'll get the value of rmin.   5.2 Potential Energy for a System of charged Particles : When more than two charged particles are there in a system, the interaction energy can be given by sum of interaction energy of all the pairs of particles. For example if a system of three particles having charges q1, q2 and q3 is given as shown in figure. The total interaction energy of this system can be given as Derivation for a system of point charges : (i) Keep all the charges at infinity. Now bring the charges one by one to its corresponding position and find work required. PE of the system is algebric sum of all the works.  Let W1 = work done in bringing first charge W2 = work done in bringing second charge against force due to 1st charge W3 = work done in bringing third charge against force due to 1st and 2nd charge. PE = W1 +  W2 +  W3  .................. (This will contain  = nC2 terms) (ii) Method of calculation (to be used in problems) U = sum of the interaction energies of the charges. = (U12 +  U13  ........ U1n) (U23 +  U24  .............. U2n) (U34 +  U35  .........U3n) ........ (iii) Method of calculation useful for symmetrical point charge systems. Find PE of each charge due to rest of the charges. If U1 = PE of first charge due to all other charges. = (U12 +  U13  ......... U1n) U2 = PE of second charges due to all other charges. = (U21 +  U23  .......... U2n) U = PE of the system =    6. Electric Potential : Electric potential is a scalar property of every point in the region of electric field. At a point in electric field, electric potential is defined as the interaction energy of a unit positive charge. If at a point in electric field a charge q0 has potential energy U, then electric potential at that point can be given as V = U/qo   joule/coulomb As potential energy of a charge in electric field is defined as work done in bringing the charge from infinity to the given point in electric field. Similarly we can define electric potential as "work done in bringing a unit positive charge from infinity to the given point against the electric forces."   Properties : (i) Potential is a scalar quantity, its value may be positive, negative or zero. (ii) S.I. Unit of potential is volt =  and its dimensional formula is [M1L2T-3I-1 ]. (iii) Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point. (iv) Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinity. (taking V¥ =0) (v) Potential decreases in the direction of electric field. (a) Electric Potential due to a Point Charge in its Surrounding : We know the region surrounding a charge is electric field. Thus we can also define electric potential in the surrounding of a point charge. The potential at a point P at a distance x from the charge q can be given as Vp =  Where U is the potential energy of charge q0, if placed at point P, which can be given as U =  Thus potential at point P is The above result is valid only for electric potential in the surrounding of a point charge. If we wish to find electric potential in the surrounding of a charged extended body, we first find the potential due to an elemental charge dq on body by using the above result and then integrate the expression for the whole body. (b) Electric Potential due to a Charge Rod : Figure shows a charged rod of length L, uniformly charged with a charge Q. Due to this we will find electric potential at a point P at a distance r from one end of the rod shown in figure shown. For this we consider an element of width dx at a distance x from the point P. Charge on this element is The potential dV due to this element at point P can be given by using the result of a point charge as Net electric potential at point P can be given as   (c) Electric Potential due to a Charged Ring : Case I : At its centre To find potential at the centre C of the ring, we first find potential dV at centre due to an elemental charge dq on ring which is given as As all dq's of the ring are situated at same distance R from the ring centre C, simply the potential due to all is added as being a scalar quantity, we can directly say that the electric potential at ring centre is . Here we can also state that even if charge Q is non-uniformly distributed on ring, the electric potential at C will remain same. Case II : At a Point on Axis of Ring If we wish to find the electric potential at a point P on the axis of ring as shown, we can directly state the result as here also all points of ring are at same distance  from the point P, thus the potential at P can be given as   Graph   (d) Electric Potential due to a Uniformly Charged Disc : Figure shows a uniformly charged disc of radius R with surface charge density s coul/m2. To find electric potential at point P we consider an elemental ring of radius y and width dy, charge on this elemental ring is dq = s. 2py dy Due to this ring, the electric potential at point P can be given as Net electric potential at point P due to whole disc can be given as (e) Electric potential due to a closed disc at a point on the edge Let us calculate the potential at the edge of a thin disc of radius 'R' carrying a uniformly distributed charge with surface density s. Let AB be a diameter and A be a point where the potential is to be calculated. From A as centre, we draw two arcs of radii r and r dr as shown. The infinitesimal region between these two arcs is an element whose area is dA = (2rq) dr, where 2qis the angle subtended by this element PQ at the point A. Potential at A due to the element PQ is From D APB, we have r = 2R cos q or, dr = - 2R sin q dq Hence   7. Relation between Electric Field Intensity and Electric Potential :   (a) For uniform electric field :   (b) Non uniform electric field (c) If electric potential and electric field depends only on one coordinate, say r :   8. ELECTRIC LINES OF FORCE The idea of electric lines of force or the electric field lines introduced by Michael Faraday is a way to visualize electrostatic field geometrically. The properties of electric lines of force are the following : (i) The electric lines of force are continous curves in an electric field starting from a positively charged body and ending on a negatively charged body. (ii) The tangent to the curve at any point gives the direction of the electric field intensity at that point. (iii) Electric lines of force never intersect since if they cross at a point, electric field intensity at the point will have two directions, which is not possible. (iv) Electric lines of force do not pass but leave or end on a charged conductor normally. Suppose the lines of force are not perpendicular to the conductor surface. In this situation, the component of electric field parallel to the surface would cause the electrons to move and hence conductor will not remain equipotential which is an absurd as in electrostatics conductor is an equipotential surface. (v) The number of electric lines of force that originate from or terminate on a charge is proportional to the magnitude of the charge. (vi) As number of lines of force per unit area normal to the area at point represents magnitude of intensity, crowded lines represent strong field while distant lines weak field. Further, if the lines of force are equidistant straight lines, the field is uniform  1. A charge particle need not follow an ELOF.  2. Electric lines of force produced by static charges do not form close loop.   9. Equipotential Surfaces : As shown in figure if a charge is shifted from a point A to B on a surface. M which is perpendicular to the direction of electric field, the work done in shifting will obviously, be zero as electric force is normal to the direction of displacement. As no work is done in moving from A to B, we can say that A and B are at same potetials or we can say that all the points of surface M are at same potential or here we call surface M as equipotential surface. Following figures show equipotential surfaces in the surrounding of point charge and a long charged wire Every surface in electric field in which at every point direction of electric field is normal to the surface can be regarded as equipotential surface. Figure shows two equipotential surfaces in a uniform electric field E. If we wish to find the potential difference between two points A and B shown in figure, we simply find the potential difference between the two equipotential surfaces on which the points lie, given as VA - VB = Ed Figure shows a line charge with linear charge density l coul/m. Here we wish to find potential difference between two points X and Y which lie on equipotential surfaces M1 & M2. To find the potential difference between these surfaces, we consider a point P at a distance x from wire as shown. The electric field at point P is 10. Electric Dipole :   A system of two equal and opposite charges separated by a small distance is called electric dipole, shown in figure. Every dipole has a characteristic property called dipole moment. It is defined as the product of magnitude of either charge and the separation between the charges is given as. Dipole moment is a vector quantity and convensionally its direction is given from negative pole to positive pole. (a) Electric field due to a Dipole (1) At an axial point Figure shows an electric dipole placed on x-axis at origin. Here we wish to find the electric field at point P having coordinates (r, o) (where r >> 2a). Due to positive charge of dipole electric field at P is in outward direction & due to negative charge it is in inward direction. (2) At an equatorial point. Again we consider the dipole placed along the x-axis & we wish to find, electric field at point P which is situated equatorially at a distance r (where r >> 2a) from origin. (b) Electric field at a general Point due to a dipole Figure shows a electric dipole place on x-axis at origin & we wish to find out the electric field at point P with coordinate (r, q)       (c) Electric potential due to a dipole. Basic Torque Concept 11.1 Potential Energy of a Dipole in Uniform Electric Field 11.2 Stable and Unstable equilibrium of a Dipole in Electric Field : We've discussed that when a dipole in an electric field E, the potential energy of dipole can be given as We also know that the net torque on a dipole in electric field can be given as It shows that net torque on dipole in electric field is zero in two situations when θ = 0o and θ = 180o as shown in figure We can see that when θ = 0o as shown in figure(a) when torque on dipole is zero, the dipole is in equilibrium. We can verify that here equilibrium is stable. If we slightly tilt the dipole from its equilibrium position in anticlockwise direction as shown by dotted position. The dipole experiences a clockwise torque which tend the dipole to rotate back to its equilibrium position. This shows that at θ= 0o, dipole is in stable equilibrium. We can also find the potential energy of dipole at θ = 0o, it can by given as  (minimum) Here at , potential energy of dipole in electric field is minimum which favours the position of stable equilibrium. Similarly when q = 180º, net torque on dipole is zero and potential energy of dipole in this state is given as  (maximum) Thus at q = 180º, dipole is in unstable equilibrium. This can also be shown by figure(b). From equilibrium position if dipole is slightly displaced in anticlockwise direction, we can see that torque on dipole also acts in anticlockise direction away from equilirbium position. Thus here dipole is in unstable equilibrium.  



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