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Example: Solve: Analysis If the electric field intensity between negative and positive plates is 200~, let us find the potential field Find the equipotential at 15.00 cm , 20.00 cm, 25.00 cm as intervals of 5.00 cm and 10.00 cm The distance y is measured from the negative plate, therefore the V15 = Ey = potential at the negative plate becomes Vo = Ey = (200 (0) = 0 V20 = Ey = Vs = Ey = (200 IGETOZIO (0.05 m) = 20.0 -or 10.0 V V25 = Ey = V10 = Ey = (200 (0.10 m) = 20.0 V At any 5.00 cm between the two plates is an The equipotential lines are everywhere perpendicular equipotential line or a line of equal or constant to the electric field lines. potential. If we drew this figure in three dimensions, it would be an equipotential surface.Example Solve: Analysis Determining the electric field from the potential A voltmeter reads 400 V across two charged, parallel difference. The potential difference between two plates, that are 0.020 m apart. What is the electric plates of a parallel plate capacitor is 400 V. If the field between them? plate separation is 1.00 mm, what is the magnitude of AV the electric field intensity between the plates? E = Ay Solution: AV (400 V) E = = 4.00 x 105 V Ay (1.00 x 10-3 m) mk y=_q r This equation gives the potential at any point that is at a distance r from a point charge q. The subscript has been dropped to make it more general. Work must be done to bring positive charges at different positions (r). This will consist of a family of concentric circles around the point charge. Each of the circles is an equipotential line. Note that the electric eld is everywhere perpendicular to the equipotential lines. In three dimensions, note that there would be a family of equipotential spheres surrounding the point charge. Example Solve: Analysis Find the equipotential eld for a positive charge of V _ k_q _ 2.0 nC at r =10.0, 20.0, 30.0, 40.0, and 50.0 cm. 3 _ r3 _ 2 kq (9.00x 109 N51; )(2.00 x 10-9C) Nm2 _ V4 = r = kq (9.00 x 109 C2 )(2.00 x 10 9C) 4 kq V5 E Compute the electric field of a point charge Solve: Analysis 2.0 nC or 2.0 x 10-9C at Ar = 30.0 cm using E = What is the electric field strength at a point that is Kq 9.00 x 109 Nm2 C2 (2.0 x 10-9C) V Ar = 0.30 m to the right of a small sphere with a E = = 200 7-2 (0.30 m)2 m charge of -4.0 x 10-C using E = Ka Kq E = = E toward the sphere or to the leftWork done in moving a charge from innity. The work done is simply the charge q multiplied by the potential V: 5.18x104VatA. How much work is required to bring a charge of q = Solve: Analysis 3.50 pic from innity to point A where V = 5.18 x 10\" V. How much work is required to bring a charge of (11 = 2.00 ,uC, q2 = 6.00 pic, (13 = 8.00 [1C from W = qV = (3.50 x 10'6C)(5.18 x 104 V) innity to point A where V = 5.18 x 104 V. w = (3.50 x 10-60 (5.18 x 104 l) = 0.181] w = (11/ = (2.00 x 10'6C)(5.18 x 104 V) c w = (11/ = (6.00 x 10-6C)(5.18x 104V) w = 4511/ = (8.00 x 10-6C)(5.18 x 104 V) W = qV 1.00 eV = (1.60 x 10-19C) (1.00 V) 1.00 eV = (1.60 x 10-19C) 1.00- 1.00 eV = 1.60 x 10-19J The eV is too small a unit for nuclear physics. Where its multiples the MeV (106 eV) and the GeV (109 eV) are commonly used. The and G respectively, signify mega (106 ) and giga (109) like in megawatts or gigawatts are used in connection with other units as well. A typical quantity expressed in MeV is the energy liberated when the nucleus of a uranium splits into two parts. Such fission of a uranium nucleus releases an average of 200 MeV; this is the process that powers nuclear reactors and weapons. What is the speed of a neutron whose kinetic energy is 50.0 ev? KE = (50.0 ev) (1.60 x10-19 ev = 8. 00 x 10-18] Since KE = - mv?and the neutron mass is 1.67 x 10-27kg, the speed of the neutron is 1 2KE 2(8.00 x 10-18]) KE = = mv2; v = (16.00 x 10-18]) m (1.67 x 10-27 kg) (1.67 x 10-27 kg) m Nm kg 2 - .m m2 V = 9.58 x 109 m = 9.58 x 109 = 9.58 x 109 kg $2 = 97, 881.7- = 9.8 x104 kgSolve: Solve: The electric field intensity between two large, charged, Two charged parallel plates are separated by a distance of parallel metal plates is 6,000-. The plates are 0.05 m 2.00 cm. If the potential difference between the plates is apart. What is the electric potential difference between 300 V, what is the value of the electric field between the them? plates? AV V = Ey = E = Ay =Solve: Solve: The potential difference between two terminals of a What is the kinetic energy in electron volts of an electron battery is 12.0 V. How much work is done by the battery whose speed is 1.0 x 106 E? in transferring 200 C of charge from one terminal to the S Given other. q = 1.60 X 1019C me = 9.1093837 x 10-31kg W = (W = 6 m 1: = 1.0 x 10 5 KB 1 2 _ Emu _ KE = (eV) (1.60 x10-19 elV) 60f8