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Transcript
13. Fundamentals of electrostatics The origins of electromagnetism go back to ancient times: IV cent. BC – Thales of Miletus (rubbed amber attracts bits of straw) Essential development of the science of electromagnetism: XVIII cent. – B. Franklin introduces two kinds of electric charges (positive on the rubbed glass and negative on the plastic rod) - C. Coulomb discovers the law of interaction between charged particles q q (Coulomb’s law) F~ 1 2 r2 XIX cent. - M. Faraday (important empirical discoverythe law of induction) - H. Oersted, A. Ampere (fundamental relations between electricity and magnetism) - J.C. Maxwell (basic laws of electromagnetism) - H. Hertz (discovers electromagnetic waves) Recent theory of electromagnetism assumes the hypothesis of charge conservation (the total charge of an isolated system, i.e. the algebraic sum of positive and negative charges is constant). γ e+ + e- (pair production ) From HRW 3 Charges of the same sign repel each other (a) and of opposite signs attract each other (b). 1 Positron annihilation e+ + e− 2γ PET (Positron Emission Tomography) The system detects pairs of gamma rays emitted indirectly by a positron-emitting radionuclide (tracer), which is introduced into the body on a biologically active molecule. Three-dimensional images of tracer concentration within the body are then constructed by computer analysis. Schema of a PET acquisition process 2 13.1. Electric field Instead of considering interaction at a distance between particles 1 and 2, one introduces a concept of an electric field: charged particle 1 sets up an electric field in the surrounding space and this field acts with a force on charged particle 2 placed in this space: charge q generates electric field E E exerts a force F on q 1 The electric field F E q0 E 2 is defined through a force acting on a positive test charge q0 (charge q0 is small enough and does not alter the electric field) (13.1) Calculate the electric field generated by two point charges: a positive q1 and a negative q2 at the point where the test charge q0 is placed. The net force is a vector sum of forces exerted by unit vectors q1 and q2 on q0: q1q0 q2q0A F F1 F2 k 2 r01 k 2 r02 r1 r2 F E1 E2 q0 F E q0 For N point charges (sources) the electric field at (x,y,z) is qi E x , y ,z k 2 r0i ri i 1 N k 1 40 in vacuum in SI units (13.2) 3 Electric field of a dipole A system of two equal charges of opposite signs and separated by a distance d is called a dipole. Calculate the electric field on the dipole symmetry axis perpendicular to the line connecting both charges. From the superposition principle + - E E1 E2 For a point charge q E1 E2 k 2 a r2 The magnitude of E is a length of the rombus diagonal E 2E1 cos 2k q a r2 2 a a r 2 2 2kqa a 2 r2 3 2 Electric field generated by an electric dipole, visualized by electric field lines. At any point the field vector is tangent to the field line. For the case r >> a one obtains Ek 2qa kp 1 p r3 r 3 40 r 3 +q/2 +q/2 where p = 2aq is a dipole moment In measurements of E we cannot determine a and q separately but only their product p. Real dipoles are not formed by systems of point charges but exhibit the properties of ideal dipoles. An example is the water molecule with a permanent dipole moment. -q The water molecule as a dipole. The valence electrons remain 4 closer to the oxygen atom 13.2. Forces acting in an electric field The force acting on a point charge q in an electric field is equal (13.3) F qE The charge (q<0) in a uniform electric field generated by two plates with charges of opposite signs A dipole in an electric field A dipole of moment p is placed in a uniform electric field E. The net torque, trying to rotate a dipole about point O along the field lines, is equal to the product of a force and a distance between the forces Fd F 2a sin 2q Ea sin pE sin (13.4) Eq.(13.4) can be written in a more general vector form (13.5) p E During rotation of a dipole the work dW is done by the field A which can be related to a change in the potential energy dEp dW dEp dW ' dW’ - a work done against field forces The potential energy of a dipole at any angle θ is 2 2 θ – angle between a dipole moment and an electric field Ep dW d pE sin d pE cos pE cos p E 2 (13.6) In eq.(13.6) the reference energy Ep=0 was chosen for θ=π/2 (Epmin = -pE). 5 13.3. Gauss’ law The flux of a vector field will be defined first. As an example we take the velocity vector v. In this case the flux is a volume rate flowing through the loop 0 vS vS cos Introducing the area vector S (perpendicular to the surface with a magnitude equal to an area of the loop) one can express the flux as v S (13.7) The flux of an electric field will be defined in an analogous way. For a non-uniform field the elementary flux is defined first d E d S The flux through the closed surface (called a Gaussian surface) is E d S (13.8) (the loop on the integral sign means the integration over the closed surface) Eq.(13.7) indicates that the flux through a Gaussian surface is proportional to the 6 net number of field lines passing through that surface. Gauss’ law, cont. The meaning of a net number of field lines is illustrated in the figure below E d S 0 the field lines are outward 0 the numbers of inward and outword lines are equal 0 qenc 0 Gauss’ law (13.9) The flux of electric field through any closed surface is equal to the net charge q enclosed by this surface the field lines are inward The Gauss’ law holds for all fields proportional to 1/r2, e.g. for the gravitational field. If we know qenc one can calculate E and vice versa. Eq.(13.9) indicates that the exact distribution of a charge inside the Gaussian surface is of no concern but in practice we apply the Gauss’ law for the cases with a certain degree of symmetry, where calculation of the integral in eq.(13.9) is simple. 7 Applications of Gauss’ law The field of a point charge Due to a symmetry the Gaussian surface is taken as a sphere of radius r centered on a point charge q. At any point the electric field is perpendicular to the surface, thus the angle θ between E and dS is zero. The Gauss’ law is expressed as q (13.10) E d S 0 The left side of eq.(13.10) is equal E dS cos E dS E dS ES S S (13.11) S Thus, from eq.(13.10) we have ES q 0 E 4 r 2 q 0 E q 40 r 2 1 (13.12) The force acting on point charge q0 in the field (13.12) is F q0 E 1 qq0 40 r 2 (13.13) Eq.(13.13) is a Coulomb’s law. Thus, one can derive the Coulomb’s law from the Gauss’ law. Gauss’ law is one of the fundamental laws of electromagnetism. 8 Applications of Gauss’ law, cont. The field of a conducting sphere As the conducting sphere is charged, the excess charge Q quickly distributes itself moving to the surface and the internal electric field becomes zero. This agrees with the Gauss’ law. For 0 r R E d S 0 E0 For r R the situation is analogous to the case of a point charge, what gives E 1 Q 40 r 2 Zeroing of an electric field inside a conductor is called screening. Electric field of a uniformly charged nonconducting plane The nonconducting thin sheet is charged on one side with a uniform surface charge density σ. Due to the symmetry of electric field we choose the Gaussian surface as a closed cylindrical surface passing through the sheet. Electric field is perpendicular to the end caps of a cylinder and tangent to the lateral curved surface. From the Gauss’ law we have E d S EA EA 0 2EA A E 9 0 2 0 Problem 1 A small nonconduting ball of mass m and charge q hangs from an insulating thread that makes an angle a with a vertical, uniformly charged nonconducting sheet of large dimensions. Taking into account the gravitational force and the force acting on the ball in an electric field, calculate the surface charge density of the sheet. Electric field produced by the sheet is equal E 2 0 and the electric force acting on charge q is equal qE. The vector sum of all forces acting on charge q, i.e. gravitational, electric and reaction forces, is zero. From the figure it can then be inferred that qE tan a mg mg E tan a q Substituting for E one can calculate the charge density mg tan a 2 0 q 2mg 0 tan a q 10 13.4. Electric potential The electrostatic field is conservative and a potential energy can be associated with it. This simplifies the calculations of work done by an electric field as the principle of conservation of mechanical energy can be applied. Following the discussion in Section 5 one can write the expression for the change in potential energy dEp in the field of electric forces as q0 (13.14) dE dW F d l q E d l p 0 The elementary change in electric potential, independent of the test charge q0, is defined as dEp (13.15) dV E d l q0 The total change in electric potential between arbitrary points 1 and 2 is obtained by integrating (13.15) 2 Q Mooving q0 along the path ACB we do the same work as in mooving along the field line between A and B. (13.16) V2 V1 E d l 1 Choosing arbitrarily the starting reference point in infinity where V1 = 0, one can write xyz V x , y ,z E d l (13.17) The potential depends only on the coordinates of a given point and electric field11 E. Electric potential of a point charge According to eq.(13.3) the potential difference in a radial field of a point charge is (as the path is radial dl dr ) B 1 Q^ r d r 2 4 r 0 A VB VA dr Q 1 B Q 1 1 VB VA 40 rA r 2 40 r rA 40 rB rA Q rB r With a reference potential equal to zero (for rB one obtains VA V Q 1 40 r ∞) (13.18) The positive point charge Q produces a radial electric field r̂ - a unit vector E Q r̂ 40 r 2 From (13.5) it follows that for r = const V= const, what determines the co called equipotential surface. For a point charge the equipotential surfaces are spheres. Cross sections of equipotential surfaces (dashed lines) shown A together with respective field lines for the fields generated by a point charge (left figure) and by an electric dipole (right figure). Equipotential surfaces are always perpendicular to electric field lines. From HRW 3 12 13.5. Capacitance Two conductors isolated electrically from each other form a capacitor; when charged, the charges on the conductors (plates) are equal in magnitude but opposite signs. The potential difference between the plates, called a voltage U, and the charge q are proportional to each other q CV1 V2 CU (13.19) The charges on the capacitor plates create an electric field in The proportionality constant C is called a capacitance the surrounding space. and its value depends on the capacitor geometry and the type of dielectric between the capacitor plates. A parallel plate capacitor This capacitor consists of two parallel plates of area A each, separated by a small distance d. The electric field between the plates is uniform (neglecting the nonuniform edge field) and can be calculated from Gauss’ law E d S q 0 A (13.20) q is a charge enclosed by a Gaussian surface on the positive plate 13 Parallel plate capacitor, cont. Vectors E and q EA 0 dS E in eq.(13.20) are parallel, then this equation reduces to (13.21) q 0A 0 The potential difference between the plates can be calculated using eq.(13.16) (13.22) V V E d l E dl cos E dl 2 2 2 2 1 1 1 1 Substituting for E from eq.(13.21) into eq.(13.22) one obtains d (13.23) V V U dl 2 2 1 1 0 0 From the definition of C (13.19) and using (13.23) one obtains for the capacitance of a parallel plate capacitor q q q A A (13.24) C 0 U d 0 qd The positive potential is that of the upper plate hence the integration path from the plate 1 to the plate 2 in eq.(13.22) is directed against the electric field line. 0 d Inserting between the capacitor plates a dielectric, it becomes polarized in the electric field and this leads to a decreasing of U (at constant q) U' U r εr – the relative dielectric permitivity (dielectric constant) of a material In effect the capacitance increases εr times C q r q A rC 0 r U U d (13.25) 14 Connection of capacitors Sample problem Calculate the equivalent capacitance for the connections of capacitors as in the figures below; C1 = 10μF, C2 = 30μF, C = 20μF Parallel connection Each capacitor has the same potential difference V, which produces charges q1 and q2 on the capacitors q1 = C1V, q2 = C2V. The total charge q=q1 + q2=(C1 + C2)V=CeqV Ceq=40 μF Serial connection From the mechanism of charging it follows that each capacitor has the same charge q and the sum of potential differences across each capacitor equals the applied potencial difference V. V1 = q/C1, V2 = q/C2. V = V1 + V2=q(1/C1+ 1/C2)=q/Ceq Ceq = C1C2/(C1 + C2) Ceq=7.5 μF This connection is neither serial nor parallel. The analysis indicates that the potentials at points a and b are equal, so the A capacitor between these points can be disconnected. In this case Ceq= C/2 + C/2 = 20 μF 15 13.6. Energy stored in an electric field Charging of a capacitor is connected with some work which has to be done. One can say that this work is stored in a form of electric potential energy in the field between the plates. This energy can be recovered by discharging the capacitor. To transfer a charge dq’ between the plates across which the voltage U’ exists requires the work dW U' dq' q' dq' C To transfer the total charge q requires the work W which is equal to the potential q energy q' q 2 CU 2 W Ep C 0 dq' 2C 2 Charing of a capacitor by transferring a charge dq’ from the negative to the positive plate. (13.26) The density of potential energy, i.e. the energy per unit volume of a capacitor is Ep CU 2 0 A U 2 1 u 0E 2 A d 2 Ad d 2 Ad 2 (13.27) Eq.(13.27) holds not only for the capacitor but can be used in each case where an electric field exists in a space. 16