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Sinai University Faculty of Engineering Science Department of Basic science

Sinai University Faculty of Engineering Science Department of Basic science. Electric Field. Electric Fields.

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Sinai University Faculty of Engineering Science Department of Basic science

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  1. Sinai University Faculty of Engineering Science Department of Basic science

  2. Electric Field

  3. Electric Fields In order to model the electrical influence of an isolated charge, we resort to a rather fictional scenario. We first "fix" (think of it as glueing) a charge (or charges) at specific coordinates in space. This charge is called the "configuration charge".electrical field", E (r) = qc / 4 p e r 2 , in Rx direction The force felt by a "test" particle in this field is F x = qt Ex, Rx r qt q Does not influence the field It is just a probe

  4. Electric Field Lines The magnitude of the force is proportional to the density of the field lines ("flux density"), which decreases with the square of the distance.

  5. Sample Problem F=qE A proton is placed in a uniform electric field E. What must be the magnitude and direction of this field if the electrostatic force acting on the proton is just to balance its weight? Solution E = = =1.0 x l0-17 N/C, directed up E must point vertically upward to the float the (positively charged) proton, because F = q0E and q0>0. mg

  6. Sample Problem In an ionized helium atom (a helium atom in which one of the two electrons has been removed), the electron and the nucleus are separated by a distance of 26.5 pm. What is the electric field due to the nucleus at the location of the electron? Solution We use Eq. 3, with q (the charge of the nucleus) equal to +2e: = ( 8.99 x 109 N.m2/C2 ) X = 4.13x 10-12 N/C.

  7. r Sample Problem The magnitude of the average electric field normally present in the Earth's atmosphere just above the surface of the Earth is about 150 N/C, directed downward. What is the total net surface charge carried by the Earth? Assume the Earth to be a conductor. Solution its average surface charge density s= q /4pr2 s =e0E= (8.85 x 1012C2/N.m2)(-150 N/C) =-l.33 x l0-9 C/m2. E= 150 N/C q = s 4R2 , s= (-1.33 x 10-9 C/m2)( 4)(6.37 x 106 m) = -6.8 x 105C = -680 kC.

  8. Sample Problem • A charged drop of oil of radius R = 2.76 mm and density r= 920 kg/m3 is • maintained in equilibrium under the combined influence of its weight and a • down-ward uniform electric field of magnitude E = 1.65 x 106 N/C. • Calculate the magnitude and sign of the charge on the drop. Express the • result in terms of the elementary charge e. -mg +q(-E) =0 qE

  9. = -9.80 m/s2 - (-8.0x l0-19 C)(1.65 X 106 N/C) 4/3(2.76 X 10-6 m)3(920 kg/m3) (b) The drop is exposed to a radioactive source that emits electrons. Two electrons strike the drop and are captured by it, changing its charge by two units. If the electric field remains at its constant value, calculate the resulting acceleration of the drop. q' = (n + 2)(-e) = 5(- 1.6 x l0-19 C) = -8.0 x 10-19 C. q' = (n + 2)(-e) = 5(- 1.6 x l0-19 C) = -8.0 x 10-19 C. and, taking y components, we obtain -mg+ q'(-E)= ma = -9.80 m/s2 + 16.3 m/s2 = +6.5 m/s2.

  10. Monopole Ea 1/r2 Dipole Ea 1/r3 Quarupole Ea 1/r4 The Electric Dipole electric dipole moment, p,p = qd. Quadruples

  11. Ex: Sodium chloride molecule The separation between Na and Cl measured for NaCl is 0.236 nm (1 nm = l0-9 m), and so the dipole moment is expected to be p=ed=(l.60x l0-19C)(0.236x l0-9m) =3.78X l0-29C.m. The measured value is 3.00 x 10-29 C.m, indicating that the electron is not entirely removed from Na and attached to a Cl. To a certain extent, the electron is shared.

  12. E Sample Problem 9 A molecule of water vapor (H2O) has an electric dipole moment of magnitude p = 6.2 x l0-30C.m. (This large dipole moment is responsible for many of the properties that make water such an important substance, such as its ability to act as an almost universal solvent.). Figure 1-15 Representation of dipole moment of water molecule in an electric field, E

  13. For a point charge F = qE=ma A Point Charge in an Electric Field What happens when we put a charged particle in a known electric field? E + + For a dipole, there is a torquet=pE sin q

  14. (a) How far apart are the effective centers of positive and negative charge in a molecule of H2O? (b) What is the maximum torque on a molecule of H2O in a typical laboratory electric field of magnitude 1.5x104 N/C? (c) Suppose the dipole moment of a molecule of H2O is initially pointing in a direction opposite to the field. How much work is done by the electric field in rotating the molecule into alignment with the field? Solution(a)There are 10 electrons and, correspondingly, 10 positive charges in this molecule. We can write, for the magnitude of the dipole moment, p= qd = (l0e)(d), in which d is the separation we are seeking and e is the elementary charge. Thus This is about 4% of the OH bond distance in this molecule. (b) The torque, t= pE sinq is a maximum when q = 900. Substituting this value in that equation yields t= pE sinq = (6.2x10-30C.m)(1.5x104N/C)(sin 900) = 9.3x10-26 N.m (c) The work done in rotating the dipole from q0 = 1800 to q = 00 is given by W = pE(cos 00 - cos 1800 ) = 2pE= (2)(6.2x10-30C.m)(1.5x104 N/C) = 1.9x10-25J

  15. Assignment 1-Write short notes on the Electrophoreses including: a-Its structure b- Its theory of operation c- Its uses 2- Solve the following problems 1, 3, 6 Please, Hand it before next lecture

  16. Electric potential For our purposes, it will be much easier to deal with another quantity, the scalar "electrical potential", rather than the electric field. For a point charge, q: V(r) = q / 4 p e r. But since the electric field is a force per unit charge, the electric potential must be energy per unit charge. E=Force/charge N/C V=Energy/Charge J/C (Volts)

  17. + Electric Potential + Mgh W=Mgh W=qV + Drift velocity V1 VE h V2 Mg qE Drift velocity - - W=W1+W2 = qV= qV1+ qV2 V=V1+V2

  18. Sample Problem What must be the magnitude of an isolated positive point charge for the electric potential at 15 cm from the charge to be + 120 V? Solution Solving for q yields q = V 4pε0 r = (120 V)(4p)(8.9 x 10–12 C2/N.m2 )(0.15 m) = 2.0 x 10–9 C = 2.0 nC This charge is comparable to charges that can be produced by friction, such as by rubbing a balloon. V(r) = q / 4 p e r.

  19. Sample Problem What is the electric potential at the surface of a gold nucleus? The radius is 7.0 x 10-15 m, and the atomic number Z is 79. Solution The nucleus, assumed spherically symmetric, behaves electrically for external points as if it were a point charge. Thus we can use Eq. 18, which gives, with q = + 79e, This large positive potential has no effect outside a gold atom because it is compensated by an equally large negative potential from the 79 atomic electrons of gold.

  20. Vca E qo Qc Vcb Electric potential Energy L Electric Field is a conservative Field W ab = F xDx = ( - q0E)(L) = - q0EL Using the definition of potential energy difference, DU = -W

  21. (-2,1) (1,1) (0,0) Electric potential Example if in a vacuum two protons (qc1) lie in the xy plane at (x1, y1) = (-2,1) Angstroms and an electron (qc2) lies at the origin ((x2, y2) = (0,0)), the energy of an electron (q t) at (x t, y t) = (1,1) Angstroms would be U = U t c1 + U t c2(= q t (V c 1 + V c 2)) = (- e) (2 e) / (4 p e 0 3 x 10 - 10) + (- e) (e) / (4 p e 0 x 10 - 10) = 4.69 x 10 - 17 J, where the distances between charges are r=((x t - x i) 2 + (y t - y i)2) and "i" denotes either of the configuration charges. Note that chemists often compute energies in kJ/mol, which would necessitate multiplying the above answer by a conversion factor of NA / 1000.

  22. + Charge acceleration Van De Graaff Accelerator +++++++++++++++++ + Utc = E L=qt Vc U= ½ mv2

  23. Sample Problem An alpha particle (q = + 2e) in a nuclear accelerator moves from one terminal at a potential of Va = +6.5 x106 V to another at a potential of Vb = 0. (a) What is the corresponding change in the potential energy of the system? (b) Assuming the terminals and their charges do not move and that no external forces act on the system, what is the change in kinetic energy of the particle? Solution(a) we have DU = U b – U a = q (V b – V a) = (+2)(1.6 x 10–19 C)[ 0 – 6.5 x 106 V] = - 2.1 x 10–12 J. (b) If no external force acts on the system, then its mechanical energyE = U + K must remain constant. That is, DE = DU + DK = 0, and so DK = - DU = +2.l x 10-12 J. The alpha particle gains a kinetic energy of 2.1 x 10-12 J, in the same way that a particle falling in the Earth's gravitational field gains kinetic energy.

  24. Problems Solve the following problems 1, 2, 4, 6, 8, 13 Cyclotron is a device used to accelerate electric charges. Write a short account on the Cyclotron. -Theory of operation -Application

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