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Phy213: General Physics III Chapter 22 Worksheet 6/30/2017 1 Electric Fields: 1. Consider a fixed point charge of +2.0 C (q1). q1 = +2.0 C P . d = 0.1 m a. What is the magnitude and direction of the electric field at a point P, a distance of 0.1 m? Ans. b. 1 q1 ˆ 6 N ˆ E= 2 i= 1.8x10 C i 4 o d A 2nd charge (q2 = -2.0 C) is placed at point P. What is the magnitude and direction of the electric force exerted on q3? Ans. 1 q1q2 ˆ ˆ FE = 2 i = -q2E = - 3.6 N i {an attractive force} 4 d o 2. Three protons are present in the nucleus of a lithium atom, forming an equilateral triangle. The distance, r, between each pair of protons is 1.5 x 10-15 m. Lithium Nucleus r + + + a. What is the magnitude of the electric field vector due to the lithium nucleus at point P, a distance of 2.5x10-10m from the center of the nucleus? Ans. Since d>>r: E 1 3qp 10 N 2 = 6.9x10 C {radial outward from nucleus} 4 o d b. What is the magnitude and direction of the total electric force exerted on an electron positioned at point P. Ans. FE = qe E = e E = 1.1x10-8 N {inward towatd nucleus} Phy213: General Physics III Chapter 22 Worksheet 6/30/2017 2 3. Consider an electric dipole, q = 0.1 C, separated by a distance (d) of 5x10-6 m. +q + -q - P . r+ E- r- E+ a. Sketch the electric field lines for the electric field vector due to the dipole. b What is the magnitude and direction of the electric field vector due to the dipole at point P, a distance (r) of 0.5 m to the right of the dipole? q q E = E++ E- = cos ˆi - sin ˆj + - cos ˆi - sin ˆj 2 2 4 r 4 r o + o - Ans. c. Ans. - 2q 2d ˆj = -q E sin 2 2 2 4 or 2 or r2+ 2d ˆj = -p ˆj = - 3.6 10-2 3 4 or N C ˆj What is the magnitude and direction of the electric field due to the dipole at a distance of 0.5 m directly above the dipole? E 1 p ˆ -2 N ˆ 3 j = 7.2x10 C j 2 o r 4. Consider a uniform line of charge with a total charge of 5 pC and a total length of 0.5 m. dE P . dE y dq dq +++++++++++++++++++++++++++++++++++ ++++ x a. Ans. What is the magnitude of the linear charge density? = 5 pC = 1x10-11 0.5 m C m b. Sketch the field lines for the line of charge in the diagram above. c. Derive an equation for the electric field vector at point P, located 0.3 m above the line and centered between the ends. Ans. Since all but the vertical component of E cancels out: 1 dq 1 dx y ˆ dEy = d E cos = 2 cos j = 2 2 4 o r 4 o x + y x2 + y2 Solving for E: 1 2 ˆj Phy213: General Physics III Chapter 22 Worksheet 6/30/2017 3 y L dx ˆj E = dE = 3 2 2 2 4 o 0 x + y L ˆj {in general, radially outward from line} E= 1 2 2 2 4 o y L + y d. Ans. e. Ans. f. Ans. What is the magnitude of the electric field vector at point P? E = 0.44 NC {directed away from the line of charge} Determine the electric force vector exerted on an electron placed at point P. FE= qeE = eE = -7.0x10-20 N ˆj{inward toward the line} What is the acceleration of the electron located at point P? a= FE = -7.7x1010 me m s2 ˆj {inward toward the line} 5. Consider a charged ring, with total charge of 2.4 mC and radius of 0.3 m. -q a. Ans. b. Ans. P . What is the magnitude of the surface charge density? = 2.4x10-3C 0.3 m 2 = 8.5x10-3 mC2 Derive an equation for the electric field vector at point P, located 1.5 m to the right of the center of the disc. Since all but the radial component of E cancels out, establish dEcos in terms of dq: 1 dq 1 rdr x ˆ d E cos = 2 cos i= 2 2 4 o d 4 o x + r x2 + r2 Solving for E: 1 2 ˆi= 1 xrdr 4 o 2 2 x +r 3 2 ˆi Phy213: General Physics III Chapter 22 Worksheet 6/30/2017 4 x R rdr E = dE = 2 4 o 0 x + r 2 3 2 ˆi R x E= 4 o -1 x 2 +R 1 ˆi = x 1 1 4 o x x2 + R 2 2 2 0 1 2 ˆi {away from ring} Phy213: General Physics III Chapter 22 Worksheet d. 6/30/2017 5 What is the magnitude and direction of the electric field vector at point P? Ans. (8.5x10-3 mC2 )(1.5m) 1 1 E= (1.5m) 4 o (1.5m)2 + (0.3m)2 E = 1.5x106 NC ˆi {away from ring} 1 2 ˆi 5. The plates of a parallel plate capacitor have a uniform charge of +1.0 C and -1.0 C, respectively. Each plate is a square with sides of length of 0.5 m. a. What is the magnitude of the surface charge density on each plate? Ans. = 1.0C 0.5 m 2 = 4.0 C m2 b. What is the magnitude of the electric field vector inside the plates of the capacitor? Ans. Treat each plate as a charged flat disc of infinite R, i.e. x << R, then the E field due to each plate respectively is then: ˆ Eplate = i {both fields will point toward the right} 2 o The total electric field between the plates is therefore: Ebetween plates = Eplate 1 + Eplate 2 = ˆi = 4.5x1011 NC ˆi o c. What is the electric force exerted on an electron placed inside the plates of the capacitor? Ans. FE= qeE = eE = -7.2x10-8 N ˆj {to the left} d. What is the acceleration of the electron while it is in between the capacitor plates? Ans. e. Ans. a= FE = -7.9x1022 me m s2 ˆj {to the left} For a 0.2 m plate separation, calculate the final speed of an electron (initially at rest), that is placed at the negative charged plate and just reaches the positive plate. v= v2 = - 2ad ˆi = - 2 -7.9x1022 sm2 -0.2m ˆi v = -1.8x1011 ˆi {a bit unrealistic eh??} Phy213: General Physics III Chapter 22 Worksheet 6/30/2017 6 Dipole Moment of Water: 6. Water is a polar molecule that consists of 2 H atoms attached to a central oxygen atom. The H atoms are oriented at an angle of approximately 105o. Each of the H-O bonds has a dipole moment ( pOH ) associated with it and together the 2 dipole moments have a resulting net dipole moment ( pH2O ) of 6.2x10-30 C.m. The effective separation distance between the respective positive (+) and negative (-) charges in each O-H bond is 3.9x10-12 m. a. What is the magnitude of the effective dipole moment of the individual O-H bond for water? Ans. pOH = pOH1 = pOH2 = b. = 5.1x10 C m H =105O pOH = qd = d= 5.1x10-30C m = 1.3x10-18C Calculate the torque on a water molecule due to the electric field of a charged sphere (qsphere = -1.5x10-6 C) at a separation distance of 0.05 m, where the direction of the electric field is oriented at an angle of 30o to pH2O . Ans. 1 qsphere 6 N E = 2 = 5.4x10 C 4 o r E = p E = pEsin30o = 1.7x10-23 N m d. Determine the magnitude of the charge () for the individual O-H dipole. Ans. c. -10 2cos(52.5o ) H pH2O = pOH1 + pOH2 = 2qdcos(52.5o )jˆ = pH2O O O What is the potential energy of the water molecule in this electric field? Ans. U = p E = pEcos30o = 2.9x10-23 J