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Capacitance

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Published in: IIT JEE Mains
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Notes On Capacitance.

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

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  1. CAPACITORS AND CAPACITANCE CAPACITY OF AN ISOLATED CONDUCTOR When charge is given to an isolated body, its potential increases and the electric field also go on gradually increasing. In this process at some stage the electric field becomes sufficiently strong to ionize the air particles around the body as a result body is not able to store any additional charge. During the process the ratio of charge Q on the body and potential (V) on the body remains constant. This ratio is called the capacity of the body C=Q/V In Sl system, the unit of capacity is coulomb/volt and is called Farad (F) The capacity of a body is independent of the charge given to it and depends on the shape and qi7P nnlv CAPACITOR Capacitor is an arrangement of two conductors carrying charges of equal magnitude and opposite sign and separated by an insulating medium. The following points may be carefully noted. (i) The net charge on the capacitor as a whole is zero. When we say that a capacitor has a charge Q, we mean that positively charged conductor has a charge +Q and the negatively charged conductor has a charge -Q. (ii) The positively charged conductor is at a higher potential than negatively charged conductor. The potential difference V between the conductors is proportional to the magnitude of charge Q and the ratio Q/V is known as capacitance C of the capacitor. Unit of capacitance is farad (F). The capacitance is usually measured in microfarad (gF) (iii) Circuit symbol is PARALLEL PLATE CAPACITOR A parallel plate capacitor consists of two metal plates placed parallel to each other and separated by a distance 'd' that is very small as compared to the dimensions of the plates. Due to this, the non-uniformity of the electric field near the ends of the plates can be neglected and in the entire region between the plates the electric field can be taken as constant. The area of each plate is A. Let Q be the charge on each plates. Surface charge is o. Direction of electric field produced by both the plates is in same direction. Where outside the plates electric field is opposite in direction hence zero Then electric field between the plates is given by
  2. potential difference (V) between plates is given by V Q c EoA d d 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 low potential conductor as sphere of infinite radius. The potential difference between these two spheres is —o 41TEoR Hence Capacitance C = Q/V = 4TTEoR ENERGY STORED IN CHARGED CAPACITOR In order to establish a charge on the capacitor, work has to be done on the charge. This work is stored in the form of the potential energy of 'the charge. Such a potential energy is called the energy of capacitor. Suppose the charge on a parallel plate capacitor is Q. In this condition each plate of the capacitor is said to be lying in the electric field of the other plate. The magnitude of the uniform electric field produced by one plate of capacitor is = — Where o is g and A is area of plate Hence taking arbitrarily the potential on this plate as zero, that of the other plate at distance d from it will be The potential energy of the second plate will be = (potential) (charge Q on it) Potential energy stored in capacitor dQ odQ Q d Q (EoA/d) A 280 2 IQ2 OR 1 2 1 2 ENERGY DENSITY OF A CHARGED CAPACITOR Energy stored in capacitor is localized on the charges or the plates but is distributed in the field. Since in case of parallel plate capacitor, the electric field is only between the plates i.e. in a volume ( AX d), the energy density 1 CV2 1 EOA V2 2 volume Ax d 2 d Ad PE (52 as E = V/d 1 — EoE2 2
  3. FORCE BETWEEN THE PLATES OF A CAPACITOR In a capacitor as plates carry equal and opposite charges, there is a force of attraction between the plates. To calculate this force, we use the fact that the electric field is conservative and in conservative field F = -d U/dx. In case of parallel plate capacitor IQ2 But C x UE = I-ex So —1 q2 dx 2 EoA The negative sign indicates that the force is attractive DIELECTRICS AND PLOLARISTAION Non-conducting material are called dielectric. Dielectric materials are of two types (i) non- polar dielectric (ii) Polar dielectric (i) Non-polar dielectric In a non-polar molecule, the centre of the positive and negative charge coincides with each other. Hence they do not possess a permanent dipole moment. Now when it is placed in a uniform electric field, these centres are displaced in mutually opposite directions. Thus an electric dipole is induced in it or molecule is said to be polarized. If extent of electric field is not very strong, it is found that this dipole moment of molecule is proportional to external electric field Eo . F = aEo Where ct is called the polarisability of the molecule The units of a is e m N -1 (ii) Polar Molecule A polar molecule possesses a permanent dipole moment p, but such dipole moments of different molecules of the substance are randomly oriented in all possible directions and hence the resultant dipole moment of the substance becomes zero. On applying an external electric field a torque acts on every molecular dipole. Therefore, it rotates and tries to become parallel to the electric field. Thus a resultant dipole moment is produced. In this way the dielectric made up of such molecules is said to be polarised.
  4. EFFECT OF DIELECTRIC ON CAPACITANCE Capacitance of a parallel plate capacitor in vacuum is given by charge density on plates is EoA co d dielectric slab Consider a dielectric inserted between the plates of capacitor, the dielectric is polarized by the electric field, the effect is equivalent to two charged sheets with surface charge densities cp and —cp . The electric field in the dielectric will be So the potential difference across the plates is V = Ed = d For linear dielectric, we expect cp to be proportional to electric field due to plates Eo Thus (c — Up) is proportional to and can be written as Where K is a constant characteristic of the dielectric. Then we have AKEO The capacitance C, with dielectric between the plates , is then Q KEOA c = KCO d The product Keo is called the permittivity of the medium denoted by E , E = KEO Or K = €0 For vacuum K = 1 and for other dielectric medium K >1. INTRODUCTION OF A DIELECTRIC SLAB OF DILECTRIC CONSTANT K BETWEEN THE PLATES (a) When battery is disconnected Let qo, Co, Vo, Eo and I.JO represents the charge, capacity, potential difference, electric field and energy associated with charged air capacitor respectively. With the introduction of a dielectric slab of dielectric constant K between the plates and the battery disconnected. Charge remains constant, i.e., q = qo , as in an isolated system charge is conserved. Capacity increases, i.e., C = KG, as by the presence of a dielectric capacity becomes K times. (iii) Potential difference between the plates decreases,
  5. q c (iv) As Field between the plates decreases, (v) Energy stored in the capacitor decreases q — qo and C [as q = qo and C = KCOI (b) When battery remains connected ( potential held constant) (i) Potential remains constant i.e V = Vo (ii) Capacity increases i.e C = KCo (iii) Charge on the capacitor increases i.e q = Kqo (iv) Electric field remains unchanged E = Eo (v) Energy stored in the capacitor increases SERIES COMBINATION OF CAPACITORS q 4. OR q q Capacitor are said to be connected in series if charge on each individual capacitor is same. In this situation V = VI + V2 + 1/3 If C is the effective capacitance of combination then we know that V = q/C and VI = q/CI , 1/2 = q/C2 , V3 = q/C3 q c q q PARALLEL COMBINATION OF CAPACITORS When capacitors are connected in parallel, the potential difference V across each is same and the charge on Cl, C2 and C3 is different i.e qi, q2 q and q3 The total charge q is given by q = qi+ q2 + q3 potential across each capacitor is same thus qi = C2V and q3 = C3V If C is equivalent capacitance then q=CV = CIV
  6. Thus + CA/ + C3V Or C = Cl + C2 + C3 Charge on capacitor Let two capacitors are connected in parallel, let Q be the total charge , Let QI be the charge on capacitor of capacity Cl and Q2 be the charge on capacitor of capacity C2 Since both capacitor have same potential from formula V = Q/C we get QI + Q2 Q2 Since Q — — QI+Q2 Q Q2 Similarly REDISTRIBUTION OF CHARGES Cl+C2 Cl+C2 If there are two spherical conductors of radius RI and R2 at potential VI and V2 respectively Far apart from each other (so that charge on one does not affect the other). The charges on them will be QI = CIVI and = The total charge on the system is Q = QI + Q2 The capacitance C = Cl + C2 Now if they are connected through a wire, charge will flow from conductor at higher potential to lower potential till both acquires same potential let charge on first becomes qi and charge on second sphere becomes qz Since potential is same We know that capacity of sphere C - - 41TEoR . Thus
  7. But Q = (11 + q2 qi + C12 similarly VAN DE GRAFF GENERATOR Principle: Suppose there is a positive charge Q, on an insulated conducting spherical shell of radius R, as shown in figure. At the centre of this shell, there is a conducting sphere of radius r (r
  8. Metal Brush Metal brush delivering charge from source Pully Insulating belt to carry and deliver charg Insulating Support column Motor driven pully Grounded metal base As shown in the figure a spherical shell of a few meter radius, is kept on an insulated support, at a height of a few meters from the ground. A pulley is kept at the centre of the big sphere and another pulley is kept on the ground. An arrangement is made such that a non-conducting belt moves across two pulleys. Positive charges are obtained from a discharge tube and are continuously sprayed on the belt using a metallic brush (with sharp edges) near the lower pulley. This positive charge goes with the belt towards the upper pulley. There it is removed from the belt with the help of another brush and is deposited on the shell (because the potential on the shell is less than that of the belt on the pulley.) Thus a large potential difference (nearly 6 to 8 million volt) is obtained on the big spherical shell. Uses : With the help of this machine, a potential difference of a few million (1 million = 106 = ten lac) volt can be established. By suitably passing a charged particle through such a high potential difference it is accelerated (to very high velocity) and hence acquires a very high energy (- mv2). Because of such a high energy they are able to penetrate deeper into the matter. Therefore, fine structure of the matter can be studied with the help of them.

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