7-Electrostat-01- Theory

Study of the interaction between charges in which one of the charge is at rest is known as electro- statics. As electrostatic interactions are of two types attractive and repulsive there are two types of charges, conventionally named as positive and negative. All bodies are composed of atoms, which contain negative charge as electrons and positive charge as protons. In such a way that in normal condition every atom is neutral (means net positive charge equals net negative charge). CHARGING OF A BODY : A neutral body can get charged only by transfer of electrons, thus the lowest unit of free charge that may appear on a body is charge of electron whose magnitude is e. When a body gets n electrons from other body charge on it becomes – ne while charge on body loosing n electrons becomes + ne. SI unit of charge is coulomb (C). Magnitude of charge of an electron e = 1.6 x 10–19C Illustration: A copper sphere contains about 2 x 1022 atoms. The charge on the nucleus of each atom is 29e. what fraction of the electrons must be removed from the sphere to give it a charge of +2  C ? Solution:  The total number of electrons is 29 (2x1022 ) = 5.8 x 1023. Electrons removed = (2x10–6C) / (1.6x10–19C) = 1.25 x 1013, so the fraction removed = electrons removed / total number electrons = 2.16 x 10–11. Since loss or gain of electron is responsible for creating charge on a body and electron is a particle with mass, every charged body will have mass also. PROPERTIES OF CHARGES : (i) Charge is quantized multiple of charge on electron Q= ± n e. (ii) conservation of charge For a system of bodies sharing charges among themselves net charge (some of positive or negative charge) remains same. Illustration: Three metallic spheres say X, Y and Z have charges 10C, –10C, 10C respectively. X, Y, Z are brought in contact such that charge on each of A and B becomes 3C what is charge on Z. Solution: Net charge initially on X, Y and Z = (+10 –10 +10) = 10C = Final net charge on X, Y and Z = qX + qY +qZ = 3+3+qc = 10 C  qc = 4C. COULOMB’S LAW : Interaction force between two point charges is given as qi q j F = k r Where r is the separation between point charges qi and qj Where F, the force between point charges qi and q j is repulsive if charges are of same nature and attractive if charges are of opposite nature. k is a constant = 1 40 1 Where 0 is known as permittivity of free space k = VECTOR FORM OF COULOMB’S LAW : (Force on jth point charge due to ith )  4π 0 9 x 109 N m2/C2 in SI unit Y → 1 q q → → F  i j (r  r ) 40 ji ˆri and ˆrj are position vectors of the point charges respectively and → → → rji rj ri Z Illustration : Two small balls each of mass m and charge q on each of them are O suspended through two light insulating string of length l from a point. Find the expression for angle made by any of the string with vertical when under static equilibrium. q q Solution: Let angle of any string with vertical be  as shown m m F  1 e 4 q 2 2𝑙 sin 2 T sin . . . (i) for horizontal direction O T cos  = mg . . . (ii) for vertical direction  Dividing (i) by (ii) T T Fe tan  = mg FORCE ON POINT CHARGES IN ANY MEDIUM: Fe Fe mg mg If two charges are placed in any other medium force on any of them is 1 F = 4  qiqj r 2 0 r ji Where r is a constant, characteristic of the medium known as dielectric constant of the medium (often denoted as K also). But in this case also interaction force between qi and qj 1 qi q j is 40 r 2 and r do not have any role. What is happening that dielectric medium is getting polarized and force on qi or qj is not simply due to qj or qi but due to polarized charges 1 also and net force on qi or qj becomes r times. r is also known as relative permittivity of the medium. 1 Illustration : Two point charges A and B have charges respectively 2 C and 2 C with their position vectors respectively as (ˆi  ˆj  kˆ) and ( → → → . Find the force on charge at A due to B. Solution: q  1 C A 2 →  ˆi  ˆj  kˆ i j 3k) q B  2C →  ˆi  ˆj  3kˆ → FAB  1 4 q q ––→ 2 rˆAB 0 | rAB | 1 qAqB r  r  1 1 9 2 ˆ ˆ ˆ = 40 3  (9 10 )  ˆ ˆ (2i  2 j  2k) ˆ  N. 2i  2 j  2k Illustration : Two equal point charges (10–3C)are placed 1 cm apart in medium of dielectric constant K = 5 . (a) Find the interaction force between the point charges. (b) Net force on any of the charge. Solution: (a) Interaction force between point charges 1 q q 10 3 2 F = 40 1 2  9 109 r 2 10 2 2 = 9 × 107 N (b) Net force 1 q q 9 109 103 2 F  40 k 1 2  r 2 5 102 2 = 18 × 106 N SUPERPOSITION OF ELECTROSTATIC FORCES: Electrostatic force between any two charges is independent of the presence of any other charge. Net force at any charge is the vector sum of all the forces acting on the charge due to all other charges. Force on ith charge in an assembling of n charges. →  n 1 qiqj rˆ F j1 40 j i 2 ij ij Illustration : Three charges each of 20C are placed along a straight line, successive charges being 2 m apart as shown in Figure. Calculate the force on the charge on the right end. Q1 Q2 Q3 F2 2m 2m 1 Solution: F = F1 + F2 F = kQ1Q2  1 r 2 9x109  20x106 2 42  0.225N Q1 Q2 Q3 F F2 2m 2m 1 kQ Q 9x109  20x106 2 F = 2 3  2 r 2 22  0.9 N F = 0.225+0.9 = 1.125 N to the right Illustration: Four identical point charges each of magnitude q are placed at the corners of a square of side a. Find the net electrostatic force on any of the charge. A B q y Solution : C q D FD q x A FB Let the concerned charge be at C then charge at C will experience the force due to charges → → → at A, B and D. Let these forces respectively be FA , FB and FD thus forces are given as → 1 q 2 q 2  ˆi ˆj  FA  along AC =    40 → 1 AC 2 q 2 40 2a   q 2 ˆ FB  40 BC 2 along BC = 4πε ( j) 0 a 2 → FD  → 1 40 → q 2 DC2 → along DC = → q 2 4πε 0 (ˆi) a 2 Fnet  FA  FB  FD q 2 ˆ 1  ˆ 1   4πε a 2 i 2  1  j 2 2  1 2 0   →  1    q 2   1  q 2 Fnet  2  1 2    2  2 . ELECTRIC FIELD:   40a  2  40a Every charge sets up around it a space in which any other charge will experience electrostatic force. This space is known as the electric field. Electric field at a point is given as the force experienced per unit charge at that position. The charge over which field is checked is purposely taken small so that it does not affect set up of → other charges. Filed is denoted by E . Mathematically. →  F  E  lt   SI Unit of Electric field is N/C q0  q  Illustration: A particle of charge 1C and mass 1 gram is suspended in air near surface of the earth such that weight of particle is balanced by electrostatic force on particle. Find the electric field at position of the particle. Solution: Electric force is equal and opposite to weight such that → Fe  W  0  → ˆ → mg ˆ  E  q j = 104 N/C vertically upward. ELECTRIC FIELD DUE TO A POINT CHARGE: Electric field due to a point charge Q lying at origin (O), at a point P given by position vector → . Force on a charge q at P → 1 Qq ˆr 4π 0 r 2 [where rˆ represents direction, away from O] Force on unit charge at P E  F  q 1 4π 0 Q ˆr r 2 Q O r P SUPERPOSITION OF ELECTRIC FIELD : In case of many point charges field at a point is given as → → E  Ei {Where  Ei represents vector sum of the field at P due to individual point charges  For a given setup of charges field is the property of a point.  Field is a vector quantity hence it can be added vectorially only, not algebraically. Illustration(a): Four point charges q are placed at corners of a square of side a. Find the intensity of the field at the (a) Centre of the square (b) Mid point of any side. Solution(a): Let position of charges be A, B, C and D and O is centre of square D q y A x q B q x → → → → Let field at O due to A, B, C, D respectively be EA , E B , EC and ED . At O magnitude of field due to each point charge are same as OA = OB = OC = OD thus EA = EB = EC = ED = 1 q  1 q  1 q 40 OA2 40 a / 2 2 20 a 2 → 1 q  ˆi ˆj  EA     2πε0 a   → 1 q   ˆi ˆj  EB     20 a   → 1 q   ˆi ˆj  EC     2πε 0 a   → 1 q  ˆi ˆj  ED     2πε 0 a  2  → → → → → E  EA  EB  EC  ED = O D C q q y (b) x EC ED Let concerned point be P → → 1 q EA  EB  4 (AP) 2 → EA  q ˆi  0 a 2 → EB  q  0 a 2  ˆi  → → 1 q 1 q q | ED |  | EC |  4 0 DP2  4 0   5   0 5a 2  a  → q 2 ˆ      1 ˆ EC   i  j 0 5a       → q  2 ˆ  1  ˆ ED   5a2  5 i   5  j  0     → → → → → q ˆ  E  EA  EB  EC  ED    5a2 ( j) 5 IN CASE OF CONTINUOUS CHARGE DISTRIBUTION (CHARGED BODY) FIELD IS GIVEN AS → → dq ˆr E  dE =  4  r 2 When dE stands for field due to an element of the body , dq is charge on that element and → is the position vector of the concerned point with respect to considered element In component form E  dEx i  dE y j  dEz k Illustration : Find the intensity of the field due to uniformly charged ring at (a) Centre of ring. Q (b) At a point on the axis of the ring. Solution (a): x Let charge Q is distributed uniformly over a ring of radius R. Let O be the centre of ring. Linear charge density on ring l = Q 2R Let us consider the ring to be lying in X–Y place and axis of the ring is oriented along Z axis. Let us consider two small element of ring of angular width dq, diametrically opposite say 1 and 2. Charge on each element dq = l Rdq. The intensity of the field at O due to each element is say dE dq dE = 4πε 0 R 2 But in opposite directions. Net field at O due to both dE  dE1  dE 2  0 Net field at O E  dE  0 (b) Let concerned point be P. Since P lies on axis position coordinate of point p is (0, 0, z) let field at p due to element 1 be dE1 ad due to 2 be dE 2 R component of dE1 and dE 2 along axis are dE cos  (where tan  = z  cos   z 2 dE1 X  dE 2  1 dq 4 0 R 2  z2 1 Rd 4 0 R 2  z 2 Component of the field perpendicular to axis are in opposite direction of same magnitude get cancelled and only component along z-axis get added. dEP = 2 dE cos  = 2R z d 4 0 R 2  z 2 3 / 2  2R 2 EP =  dEp  2R =  4  0 z R1  z  Q 2 3 / 2  d z 4 0 R 2  z 2 3 / 2 4 0 R 2  z 2 3 / 2 . ELECTRIC LINES OF FORCE:  Electric lines of force are imaginary lines that we draw to aid our visualisation of the properties of the electric field. The tangent to these lines at any point represents the direction of field intensity.  The electric lines of forces are continuous curves in an electric field starting from a positively charged body and ending on a negatively charged body.  The tangent to the curve at any point gives the direction of the electric field intensity.  Lines of forces never intersect each other.  The lines of forces are always normal to the surface of a conductor while leaving the conductor or ending on it.  At points where the intensity is low, the lines of forces will be widely separated and where the intensity is higher, the lines of force will be closely packed. Electric lines of forces due to +ve charge Electric lines of forces due to -ve charge Electric lines of forces due to two equal +ve charges (field is zero at O) Þ O is a null point. Electric lines of forces due to two equal but oppositely charged bodies ELECTRIC DIPOLE : An arrangement of two equal and opposite charges separated by a small distance is known as an electric dipole. Let q and –q be two charges separated by distance 2l. The dipole moment of the dipole is : – P -q 2l +q → → p  q 2 𝑙 It is a vector quantity and is directed from –ve charge towards the +ve charge. The line joing -q to +q is known as the axis of the dipole. Electric field at axis : (Line joining the charges) Electric field due to a short dipole on its axis at a point A at a distance r from dipole ( 𝑙 <R). This surface encloses the entire charged sphere. So from Gauss’s law, we have E(4r 2 )  Q 0 or, E  1 Q 40 r 2 ...(iv F) The field at points outside the sphere is a same as that of a point charge at the centre. Variation of E with the distance from the centre (r) E 1 Q 40 R 2 O CONDUCTORS : A conductor contains “free” electrons, which can move freely in the material, but cannot leave it. Now, when an excess charge is given to an insulated conductor, it sets up electric field inside the conductor. The free electrons will redistribute themselves and within a fraction of a second (approx 10–12s) the internal field will vanish. Thus in electrostatic equilibrium the value of → at all points within a conductor is zero. This idea, together with the Gauss’s law can be used to prove interesting facts regarding a conductor. CAVITY INSIDE A CONDCUTOR: Consider a charge + q suspended in a cavity in a conductor. Consider a Gaussian surface just outside the cavity and inside the conductor. → on this Gaussian surface as it is inside the conductor. From Gauss’s law, net charge inside is zero. Gaussian surface This concludes that a charge of – q must reside on the metal surface of the cavity so that the sum of this induced charge – q and the original charge +q within the Gaussian surface is zero. In other words, a charge q suspended inside a cavity in a conductor induces an equal and opposite charge -q on the surface of the cavity. Further as the conductor is electrically neutral, a charge +q is induced on the outer surface of the conductor. The field lines coming from q cannot penetrate into the conductor, as shown in the above figure. The same line of approach can be used to show that the field inside the cavity of conductor is zero when no charge is kept inside it. ELECTROSTATIC SHEILDING: Suppose we have a very sensitive electronic instrument that we want to protect from external fields that might cause wrong measurements. We surround the instrument with a conducting box or we keep the instrument inside the cavity of a conductor. By doing this, the charge in the conductor is so distributed that the net electric field inside the cavity becomes zero and so instrument is protected from the external fields. This is called electrostatic shielding. FORCE ON THE SURFACE OR A CHARGED CONDUCTOR: In a charged conductor the charge resides entirely on the surface. Every element of the surface of the conductor experiences a normal outward force. This force is produced as a result of repulsion of the charge on the element by the similar charge on the rest of the surface of the conductor. Let us calculate this force. Let dS be a small element of the surface of a charged conductor. Let  be the surface density of charge. Let us consider a point P just outside the surface. The magnitude of the electric intensity at P is given by E   0 and is directed along the outward drawn normal to the element. The intensity E can be considered as made up of two parts: (i) an intersity E1 due to the charge on the element dS, and (ii) and intensity E2 due to the charge on the rest of the surface of the conductor. Since their directions are the same, we have E  E1   E 2  0 ...(i) Let us now consider a point Q just inside the surface . The intensity at Q may again be considered as made up of two parts. The intensity due to the charge on the element dS is equal and opposite to that at P i.e.,  E1 , since Q is very close to P but on the opposite side of the surface. The intensity due to the charge on the rest of the surface is same in magnitude and direction as at P i.e., E2 (since Q is very close to P). But the resultant intensity at Q must be zero, since Q line inside the conductor. Hence  E1  E 2  0 or E1  E 2 Substituting this in equation (i), we get  2E 2   0  E  σ 2ε 0 This gives the outward force experienced by a unit positive charge on the elements dS due to the charge on the rest of the surface. Since the charge on the element is  dS, the force on dS is F  E 2 ( dS)   2dS F 2 Hence the force per unit area of the surface is F   2 dS 20 ...(v A) Whatever the sign of  , this force acts outward along the normal to the surface. Now, from equation (i) E   0 , so that   0 E . Substituting this value of  in equation (v A), the outward force per unit area of the surface  (0 E)2 20  0 E 2 2 ...(v B) Hence the force per unit area (or electrostatic pressure) experienced by a charged conductor is 2 / 20 or 0 E 2 / 2 newton/meter2 directed along the outward drawn normal to the surface. ENERGY DENSITY OF AN ELECTRIC FIELD : Work is done in creating any electrostatic system. This work is stored as energy in the field. The energy per unit volume or the energy density U, of the field is given as U  1 Kε 2 0 E 2 ...(vi) ELECTRIC POTENTIAL: The electric potential at a point in an electric field is the external work needed to bring a unit positive charge, from infinity (point of zero potential) to the given point. Thus, V  Wext q 0 ..(vii) Where Wext is work done in moving a charge q0 from infinity to given point. Important points regarding electric potential (i) As electric field is conservative, Wext  U So, Or, V  U q 0 U  q0 V Thus, the electric potential at a point is numerically equal to the potential energy per unit charge at that point. (ii) It is a scalar having SI unit (J/C) called volt(V). 1V  1J 1C (iii) If VA and VB are the electric potentials of two points A and B, the potential difference between A and B is equal to VB  VA . Thus the potential difference between two points. A and B, is defined as VB  VA  WAB q where from A to B. WAB 0 is the work done by an external agent in moving a positive test charge q0 (iv) We know that V  Wext q 0 → → Now, Wext   Fext .d l Since the external force is equal and opposite of the electrostatic force, we have → → Fext  qE → → or Wext   qE. dl The figure shown a curved path in a non-uniform field. The potential difference between the point A and B is given by B → → VB  VA   E.d𝑙 ...(viii) A A Since the electrostatic field is conservative, the value of this line integral depends only on the end points A and B not on the path taken. Illuatratrion: The electric field in a region is given by →  A / x3 iˆ . Write a suitable SI unit for A. Write an expression for the potential in the region assuming the potential at infinity to be zero. Solution: The SI unit of electric field is N/C or V/m. Thus, N  m3 The unit of A is or V-m2. C (x, y, z) → → V(x, y, z) = -  (x, y, z) A dx  E.dr A . = -  x 3 2x 2 ELECTRIC POTENTIAL AT A POINT DUE TO A POINT CHARGE: → 1 q → r → → E  40 r 3 r and V  E.dr  r 1 q → → 1 r q O  V   4 r 3 r.dr   4  2 dr 0  +q r A or V  1 q , ...(ix) 40 r where r is the distance of A from the point charge q. The electric potential at A(VA ) is positive if the point charge q is positive. VA will be negative if the point charge q is negative. ELECTRIC POTENTIAL DUE TO A GROUP OF POINT CHARGES: The potential at any point due to a group of point charges is the algebraic sum of the potentials contributed at the same point by all the individual point charges. V  V1  V2  V3  .... ...(x) Illustration : Two points charge q and –2q are placed at a distance 6a apart. Find the locus of the point in the plane of charges where the field potential is zero. Solution: Let us take the charge on X-axis; q at A (0, 0) and –2q at B(6a, 0) Potential at a point P(x, y) is V  q   2q V  0 q 2  x 2  y2  4q 2 (x  6a) 2  y 2  the locus is (x  6a) 2  4x 2  3y 2 . 3x 2  3y 2  2(6a)x  36a 2  x 2  y 2  4ax  12a 2 (x  2a) 2  y 2  16a 2 A circle with centre (-2a, 0) and radius 4a. ELECTRIC POTENTIAL DUE TO A CONTINUOUS CHARGE DISTRIBUTION: The electric potential due to a continuous charge distribution is the sum of potentials of all the infini- tesimal charge elements in which the distribution may be divided V   dV V  dq 40r ELECTRIC POTENTIAL DUE TO A CHARGED RING : A charge Q is uniformly distributed over the circumference of a ring. Let us calculate the electric potential at an axial point at a distance r from the centre of the ring. The electric potential at P due to the charge element dq of the ring is given by dV  1 dq  1 dq 40 Z 40 (R 2  r 2 )1/ 2 Hence, the electric potential at P due to the uniformly charged ring is given by V   1 dq  1 1  dq 40  (R 2  r 2 )1/ 2 . 40 (R 2  r 2 )1/ 2 ELECTRIC POTENTIAL DUE TO A CHARGED DISC AT A POINT ON THE AXIS: A non-conducting disc of radius ‘R’ has a uniform surface charge density  C / m 2 . Let us calcuate the potential at a point on the axis of the disc at a distance ‘r’ from its centre. The symemtry of the disc tells us that the appropriate choice of element is a ring of radius x and thickness dx. All points on this ring are at the same distance Z  , from the point P. The charge on the ring is dq  dA  (2x dx) and so the potential due to the ring is z  dV  1 40 dq  1 Z (2 x dx) Since potential is scalar The potential due to the whole disc is given by V   20 0 x dx   20   20 (x 2  r 2 )1/ 2 R (R 2  r 2 )1/ 2  r ...(x B) Let us see this expression at large distance when r  R . V  1 40 Q r , where Q  r2 is the total charge on the disc. Thus, we conclude that at large distance, the potential due to the disc is the same as that of a point charge Q. ELECTRIC POTENTIAL DUE TO A SHELL: A shell of radius R has a charge Q uniformly distributed over its surface. Let us calcuate the potential at a point (a) outside the shell; (r > R) (b) inside the shell (r < R). Solution:(a) At points outside a uniform spherical distribution, the electric field is → 1 E  4 Q rˆ r 2 since → is radial, → →  Edr E E.dr since V()  0 , we have r Q Q  1r V(r)  V()   4  V  1 Q (r  R) dr  r 2 40  r  ...(x D) 40 r We see that the potential due to a uniformly charged shell is the same as that due to a point charge Q at the centre of the shell. (b) At an internal Point At points inside the shell, E  0 . So, the work done in bringing a unit positive charge from a point on the surface to any point inside the shell is zero. Thus, the potential has a fixed value at all points within the spherical shell and is equal to the potential at the surface. V  1 Q ...(x E) 40 R Variation of electric potential with the distance from the centre (r) E 1 Q 40 R 2 O All the above results hold for a “conducting sphere also whose charge lies entirely on the outer surface. ELECTRIC POTENTIAL DUE TO A NON-CONDUCTING CHARGED SPHERE A charge Q is uniformly distributed throughout a non-conducting spherical volume of radius R. Let us find expressions for the potential at an (a) external point (r > R); (b) internal point (r < R) where r is the distance of the point from the centre of the sphere. (a) At an external point] Let O be the centre of a non-conducting sphere of radius R, have a charge Q distributed uniformly over its entire volume. Let us divide the sphere into a large number of thin concentric shells carrying charges q1, q2 , q3, ... etc. The potential at the point P due to the shell of charge q1 is 1 q 40 r . Now, potential is a scalar quantity. Therefore the potentials V due to the whole sphere is equal to the sum of the potentials due to all the shells.  V  1 q1   q 2  .... 40 r 40 r  4 r q1  q 2  q 3  ..... But  q1  q 2  q3   Q , the charge on the sphere V  1 Q ...(x F) 40 r (b) Potential at an internal point Suppose the point P lies inside the sphere at a distance r from the centre O, if we draw a concentric sphere through the point P, the point P will be external for the solid sphere of radius r, and internal for the outer spherical shell of internal radius r and external radius R. The charge on the inner solid spheres given by 4 r3 .Therefore the potential 3 V1 at P due to this sphere is V  1 1 4 4 / 3r 3 r  r 2 30 Let us now find the potential at P due to the outer spherical shell. Let us divide this shell into a number of thin concentric shells and consider one such shell of radius x and infinitesimally small thickness dx. The volume of this shell = surface area × thickness = 4x 2 dx . The charge on this shell, dq = 4x 2 dx . The potential at P due to this shell dV  1 2 4 dq  1 x 40 4x 2 (dx) x  x dx 0 The potential V2 at P due to the whole shell of internal radius r and axternal radius R is given by R V2    R x dx  r 0  (R 2  r 2 ) 20 Since the potential is a scalar quantity, the total potential V at P is given by V  V1  V2  r 2  (R 2  r 2 ) 30 20  (3R 2  r 2 ) 60 But   Q 4 R 3 3 V  1 Q 3R 2  r 2  40 2R 3 ...(x G) CALCULATION OF ELECTRIC FIELD FORM ELECTRIC POTENTIAL: In rectangular components, the electric field is → ˆ ˆ ˆ E  E x i  E y j  Ez k ; and an infinitesimal displacement is dr  dx i  dy j  dz k Thus, dV   → → ...(xi)  [Ex dx  E y dy  Ez dz] for a displacement in the x-direction, dy  dz  0 and so dV  Ex dx. Therefore,  dV  Ex  dx   y, z constant A derivative in which all variables except one are held constant is called partial derivative and is written with  instead of d. The electric field is, therefore, → V ˆ V ˆ V ˆ E   x i  y j  z k ...(xii) EQUIPOTENTIAL SURFACES: If we join the points in an electric field, which are at same potential, the surface (or curve) obtained is known as equipotential surface (curve). Important Points Regarding Equipotential surface (i) The lines of force are always normal to equipotential surfaces (ii) The net work done in taking a charge from A to B is zero if A and B are on same equipotential surface. Example: (i) In the field of a point charge, the equipotential surfaces are spheres centered on the point charge. (ii) In a uniform electric field, the equipotential surfaces are planes which are perpendicular to the fields lines. (iii) In the fields of an infinite line charge, the equipotential surfaces are co-axial cylinders having their axes at the line charge. (iv) The surface of a conductor is an equipotential surface and the inside of conductor is equipoten- tial space. Hence there is no electric field (and charge) inside the conductor’s surface. The lines of forces are always normal to the surface of a conductor. ELECTRIC POTENTIAL ENERGY: In the figure, if a charge +q is moved from B to C in the electric field of charge +Q, the work will have to be done by some outside energy in pushing the charge +q against the force of field of +Q. +Q C B This situation is very similar to that of a mass moved in gravitational field of earth away from it. Work done against the gravitational pull of earth is stored in Gravitational potential energy and can be recovered back. Similarly in electric field, work done against an electric field is stored in the form of electric potential energy & can be recovered back. If the charge +q is taken back from C to B, the electric force will try to accelerate the charge and hence to recover the potential energy stored in the form of kinetic energy. As the work done against an electric field can be recovered back, electrostatic forces and fields fall under the category of conservative forces and fields. Another property of these fields is that the work done is independent of path taken from the one point to the another. POTENTIAL ENERGY OF A SYSTEM OF TWO POINT CHARGES: The potential energy possessed by a system of two-point charges q1 and q 2 separated by a dis- tance r is the work required to bring them to this arrangement from infinity. This electrostatic potential energy is given by U  q1q 2 40 r ...(xiii) Note : While writting potential or potential energy charges must be multiplied with there signs. ELECTRIC POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGES: The electric potential energy of such a system is the work done in assembling this system starting from infinite separation between any two-point charges. For a system of point charges q1 , q 2 , qn , the potential energy is 1 n n qi q j U  2  4πε r (i ...(xiv) i1 j1 0 ij It simply means that we have to consider all the pairs that are possible. Important points regarding Electrostatic potential energy (i) Work done required by an external agency to move a charge q from A to B in an electric field with constant speed WAB  qVB  VA  (ii) When a charge q is let free in an electric field, it loses potential energy and gains kinetic energy, if it goes from A to B, then loss in ponetial energy = gain in kinetic energy or q(V  V )  1 mV 2  1 mV 2 B A 2 B 2 A Illustration: What is work done by the electrostatic field when we put the four charges together, as shown in the figure. Each side of the square has a length a. Initially charges were at infinity. +q -q a -q a +q Solution: Ui  0 [Where charges are separated by infinite distance] 1   4q 2 q 2  Uf      4πε0  a  [for 6 pairs of charges] Work done by field  ΔU  Ui  Uf  2    4   . 4πε 0 a  a  CAPACITORS CAPACITORS : Capacitor is an arrangement of two conductors carrying charges of equal magnitudes and opposite sign and separated by an insulating medium. The following points may be carefully noted. 1. The net charge on the capacitor as a whole is zero. When we say that a capacitor has a charge Q, we mean that the positively charged conductor has charge +Q and negatively charged conductor has a charge –Q. 2. The positively charged conductor is at a higher potential than the negatively charged conductor. The potential difference V between the conductor is proportional to the charge magnitude Q and the ratio Q/V is known as capacitance C of the capacitor. C  Q V Units of capacitance are farad (F). The capacitance is usually measured in microfarad (F). 1 μF = 10-6 F 3. In a circuit, a capacitor is represented by the symbol : 4. Capacitors work as a charge – storing or energy – storing devices. A capacitor can be thought of as a device which stores energy in the form of electric field. Energy stored in a capacitor is denoted by U. If V is the potential difference across the capacitor and Q is the charge on the capacitor and C is the capacitance of capacitor, then : U  1 CV 2 1 Q 2 U  U  1 QV 2 or 2 C or 2 PARALLEL PLATE CAPACITOR : The parallel plate capacitor consist of two metal plates placed parallel to each other and separated by a distance that is very small as compared to the dimension of the plates Electric field is given by σ d E = k  – + – + Where  : surface charge density on either plate – + K : dielectric constant of the medium between plates – + + If d is the separation between plates and A is the area of each plate, – + the potential difference (V) between plates is given as : V = Ed σ V = k 0 d = Q d k 0 A C = k 0 A d for parallel plate capacitor. If there is vacuum between the plates, k = 1. Illustration: A parallel plate air capacitor is made using two plates 0.2 m square, spaced 1 cm apart. It is connected to a 50 V battery. (a) what is the capacitance ? (b) what is the charge on each plate ? (c) what is the energy stored in the capacitor ? (d) what is the electric field between the plates ? (e) if the battery is disconnected and then the plates are pulled apart to a separation of 2 cm, what are the answers to the above parts ? Solution : (a) C0  ε 0 A d0  8.85 1012  0.2  0.2 0.01 C0  3.54 105 μF (b) Q0  C0 V0  3.54 10.5  50μC  1.77 103C (c) U  1/ 2C V 2  1/ 23.54 1011 502 0 0 0 U0  4.42 108 J. E  V0  50  5000V / m. (d) d0 0.01 (e) If the battery is disconnected, the charge on the capacitor plates remains constant while the potential difference between plates can change. C  A0  1.77 105 μF 2d Q  Q0  1.77 103 μC V  Q  Q0  2V  100 volts. C C0 / 2 1 Q 2 1 Q2 8 U    0   2U 0  8.84 10 J 2 C 2 C0 / 2 E  V 2V0  E  5000 V / m. c 2d 0 work has to be done against the attraction of plates when they are separated. This gets stored in the energy of the capacitor. Illustration: A parallel plate capacitor has plates of area 4 m2 separated by a distance of 0.5 mm. The capacitor is connected across a cell of emf 100 volts. (a) find the capacitance, charge & energy stored in the capacitor. (b) a dielectric slab of thickness 0.5 mm is inserted inside this capacitor after it has been disconnected from the cell. Find the answers to part (a) if k = 3. Solution : (a) C  0 A  0 d 8.85 1012  4 0.5 103 C0  7.08 102 μF. Q0  C0 V0  7.08 102 100μC  7.08 μC U  1 C 0 2 0V2  354 106 J. (b) as the cell has been disconnected, Q = Q0 C  kε 0 A  kC d 0  0.2124 μF V  Q  C 1 Q Q0 kC0 1 V0  100 volts. k 3 Q2 U 6 U  0  0  0  118 10 J. 2 C 2 kC0 k Electric field inside the plates = E  V  V0  E0 d kD k Note that the field becomes 1/k times by insertion of dielectric. ISOLATED SPHERE AS A CAPACITOR : A conducting sphere of radius R carrying a charge Q can be treated as a capacitor with high-potential conductor as the sphere itself and the low.-potential conductor as a sphere of infinite radius. The po- tential difference between these two sphere is: Q V   0 4π 0 R Q Capacitance (C) = V C=40R CYLINDRICAL CAPACITOR : Cylindrical capacitor consist of two co-axial cylinders of radii a and b and length l. The electric fields exists in the region between the cylinders. Let K be the dielectric constant of the material between the cylinders. The capacitance is given by: l 2πK 0 l C  log b a SPHERICAL CAPACITOR : A spherical capacitor consist of two concentric spheres of radii a and b as shown. The inner sphere is positively charged to potential V and outer sphere is at zero potential. The inner surface of the outer sphere has an equal negative charge. The potential difference between the spheres is : V – 0 = Q  Q 4π 0 a Q Capacitance = C = V 4π 0 b 4π 0 ab C = b  a For a dielectric (K) between the spheres : C = 4πK 0 ab . b  a CAPACITORS IN SERIESAND PARALLEL COMBINATION : SERIES COMBINATIONS : When two or more than two capacitors are connected in such a way that plates of capacitors are conneted with each other the combination is known as series. [Only first plate of first capactiors and second plate of last capacitor is connected to source. When capacitors are connected in series, the magnitude of charge Q on each capacitor is same. The potential difference across C1 and C2 is different i.e., V1 and V2. Q = C1 V1 = C2 V2 The total potential difference across combination is : V = V1 + V2 C1 + – + – V = Q  Q V  1  1 V1 V2 C1 C2 Q C1 C 2 The ratio Q/V is called as the equivalent capacitance C between point a and b. The equivalent capacitance C is given by : 1  1  1 C C1 C2 The potential difference across C1 and C2 is V1 and V2 respectively, given as follows : V  C2 V & V  C1 V 1 C  C 2 C  C 1 2 1 2 In case of more than two capacitors, the relation is : 1  1  1 C C1 C2  1  1 C3 C4  ......... 1 n 1  C C . eq i1 i Illustration: Two capacitors of capacitance C1 cell of emf 18 V. Calculate : (a) the equivalent capacitance = 6 μ F and C2 = 3  F are connected in series across a (b) the potential difference across each capacitor (c) the charge on each capacitor. Solution : (a) 1  1  1 C  C  C1 C2 C1C2 C1  C2  6  3  2μF. 6  3 (b) V1  C2 V  C1  C2 3 6  3 18  6 volts V2  C1 C1  C2 V  6 6  3 18  12 volts Note that the smaller capacitor C2 has a larger potential difference across it. (c) Q1  Q 2  C1V1  C2 V2  CV charge on each capacitor = Ceq V  2μF18 volts  36μC PARALLEL COMBINATIONS : When two or more than two capacitors are connected in such a way that one plate of all capacitors are connected to one point and other plate of all capacitors are con- nected to other single point such a combination arragement of capacitors is known as perallel combination. – C1 When capacitors are connected in parallel, the potential difference V across each is same and the charge on C1, a b C2 is different i.e., Q1 and Q2. The total charge is Q given as : Q = Q1 + Q2 Q  C1V  C2V Q V = C1 + C2 Equivalent capacitance between a and b is : C = C1 + C2 The charges on capacitors is given as : Q1 = Q2 = C1 Q C1  C 2 C2 Q C1  C 2 In case of more than two capacitors, C = C1 + C2 + C3 + C4 + C5 +………. Illuatration: Find the equivalent capacitance of the combination shown in figure between the points P and N. P 10F Q 30F N 20F Solution: The 10 F and 20 F capacitors are connected in parallel. Their equivalent capacitance is 10 F  20 F  30 F . We can replace the 10 F and the 20 F capacitors by a single capacitor of capacitance 30 F between P and Q. This is connected in series with the given 30 F capacitor. The equivalent capacitance C of this combination is given by 1  1  1 or, C  15 F . C 30F 30F CAPACITORS WITH MORE THAN ONE DIELECTRIC SLABS : (I) A parallel plate capacitor contains two dielectric slabs of thickness d1, d2 and dielectric constants k1 & k2 respectively. The area of the capacitor plates and slabs is equal to A. Considering the capacitor as a combination of two capacitors in series, the equivalent capacitance C is given by : 1  1  1 C C1 C2 1 1  d1 C k1 0 A 0 A  d 2 d k 2 0 A C = d1  d 2 k1 k 2 In general for more than one dielectric slab : 0 A C = di k i If V is the potential difference across the plates, the electric fields in the dielectrics are given as : V1  Q  CV  k1 0 A A0  E1 = d C d C d use C1  d and C = d d  1 1 1 1 1 1  V  1 1 + 2 K1 K2 1  V    d d k   d d k E1 = 1  2  1  E2 = 1  2  2  k1 k 2 k1 k 2 Note : k1 E1 = k2 E2 and E1 d1 + E2 d2 = V. (II) If there exists a dielectric slab of thickness t inside a capacitor whose plates are separated by distance d, the equivalent capacitor is given as : C = t K 0 A  d  t 1 (K = 1 for vacuum) d t C = t 0 A  d  t K The equivalent capacitance is not affected by changing the distance of slab from the parallel plates. If the slab is of metal, the equivalent capacitance is : 0 A C = d  t (for a metal, K =  ) . (III) Consider a capacitor with two dielectric slabs of same thickness d placed inside it as shown. The slabs have dielectric constants k1 & k2 and areas A1 & A2 respectively. Treating the combination as two capacitors in parallel, C = C1 + C2 k1 0 A1  k 2 0 A 2 C = d d d C = 0 [k A  k A ] . d 1 1 2 2 A Illustration: The plates of the capacitor formed by inserting four-di- electric slabs (as shown) have an area equal to S. Find the equivalent capacitance between A and B if d/2 K1 = 2 K2 = K3 = K4 = 5. Solution : Consider the capacitor as a parallel combination of C1 and C2 and series combination of C3 and C4 Equ. Capacitance = C1 + C2 +C3C4/C3 + C4 d/2 ε 0 k S  k S   ε 0S/ 2 d  1 4 2 4  d / 2 d / 2    k 3 k 4 A  ε 0S 5 ε0S  5  4d  ε 0S k d  2   k  ε0S  k 3k 4  4d 1 2    3 4   15ε 0S . 4d FORCE ON ANY PLATE OF PARALLEL PLATE CAPACITOR DUE TO OTHER: Intensity of the field at surface of any plate due to other is half of the field between plates E = 2 =  2 0   E 2ds -Q +Q Force for area dS on any plate dF = 2 A Net force on any plate F  dF  0 ds  2 2 0 – + – + – + – + – + – + – + – + Force per unit area  2 2 0 The energy stored in a capacitor is electrostatic potential energy. When we pull the plates of a capacitor apart, we have to do work against the electrostatic attraction between the plates. When we Q2 increase the separation between the plates from d1 to d 2 , an amount 2A0 (d 2  d1 ) of work is performed by us and this much energy goes into the capacitor. On the other hand, new electric field Q2 is created in a volume A(d 2  d1 ) . We conclude that the energy 2A0 (d 2  d1 ) is stored in the volume A(d 2  d1 ) which is now filled with the electric field. Thus, an electric field has energy associated with it. The energy stored Q 2 (d 2  d1 ) u  2A0 A(d 2  d1 )  Q2 2A 20   0    0 E 2 2  A0  2 where E is the intensity of the electric field . Once it is established that a region containing electric field E has energy 1  2 0 E 2 per unit volume, the result can be used for any electric field whether it is due to a capacitor or otherwise. Illustration: Plates of a parallel plate of area A and separtion between the plates d. Is charged to a potential difference of V. Find the attraction force between plates Q  CV  0 A V Solution:    d  Q2 (0 A / d)2 V2  AV2 2 0 A 2 0 A  0 . 2d2 FORCE ON A DIELECTRIC PARTIALLY INTRODUCED BETWEEN THE PARALLEL PLATES : C   (K 1)x  Where C0 1   L  C  0 A 0 d d dC  C dx 0 (K 1) L . (not connected to battery) Q -Q Capacitors of a capacitiors with partially introduced dielec- tric between the plates. x Q2 U = 2C F =  dU dx   Q 2 dC 2C2 dx Q 2 K  1C0  = 2C2  L  +ve sign mean force is inward. Since C changes with x F is variable