Week 3 Electric Potential )الشرح واألفكار الرئيسية( النوتات األساسية تتكون النوت من عشرة أسابيع. يحتوي كل نوت على شرح وحلول ألمثلة وتمارين من هوموركات وامتحانات سابقة. نوتات األمثلة والتمارين اإلضافية يتوفر على الموقع نوتات األسابيع لألمثلة والتمارين اإلضافية من واقع هوموركات وامتحانات سابقة. حقوق النوت لتعميم الفائدة فإني أتيح حقوق النسخ وإعادة اإلصدار وشروحات الفيديو لكل البشر سواء لبعض األجزاء أو الكل. واتساب )00965-94444260( يتم تنقحة النوت وتحديثها وإضافة الجديد إليها بشكل دوري. نتشوق النتقادك الهادف. العقبات هي تلك األشياء المخيفة التي تراها عندما ترفع نظرك عن هدفك. لطلب نسخة ملونة مجانا 94444206 )1(
ELECTRIC POTENTIAL ENERGY موجبة في اتجاه مجال منتظم فإنها تتأثر بقوة ) o (F = E q لها (E) إذا وجدت شحنة ) o q) نفس اتجاه المجال. في نفس اتجاه القوة (F) مسافة (d) فإن الشغل المبذول بواسطة هذه U = E q o d إذا تحركت الشحنة ) o q) الشحنة : طاقة وضع الشحنة تقل طالما تحركت في نفس اتجاه تأثير أي كتلة تتحرك في نفس اتجاه الجاذبية, و العكس صحيح. القوة (F) تماما كما تقل طاقة وضع 23.3 A positive charge moving (a) in the direction of electric field E and (b) in the direction opposite E. قد ال تكون تعيسا كما تظن. )2(
في حال شحنة ) o q) سالبة فإن طاقة وضع الشحنة يزداد القوة, والعكس صحيح. إذا تحركت في نفس اتجاه تأثير (q) في حال نشأة المجال الكهربي الذي يؤثر علي الشحنة إعادة صياغة قانون ال (U) كما يلي: ) o q )عن شحنة أخرى فإنه يمكن U = E q o d = (k q r 2) (q o) (r) = k q q o r الجز عند المصيبة, مصيبة أخرى. )3(
Potential (Voltage) potential energy per unit charge. تؤثر علي كل وحده شحنات من ( o q(. V = U q o = (k qq o r )/q o = k q r هي مقدار الطاقة التي Kinetic Energy: K = 1 2 m υ2 Coservation of energy: K a + U a = K b + U b Summation of Electric Potentual: في حال عدة شحنات فإنه يتم جمع تأثيرها في حال حساب (U), (v). U = k q o Σ q r V = k Σ q r األماني رءوس 2 أموال المفلسين )4(
A point charge q 1 = +2.40 μc is held stationary at the origin. A second point charge q 2 = 4.30 μc moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0.250 m. How much work is done by the electric force on q 2? w a b = U a U b U a = 1 4πε ο q 1 q 2 r a = (8.988 10 9 ) (2.40 10 6 )( 4.30 10 6 ) 0.150 U a = 0.6184 J U b = 1 4πε ο q 1 q 2 r b = (8.988 10 9 ) (2.40 10 6 )( 4.30 10 6 ) 0.3536 U b = 0.2623 J w a b = U a U b = 0.6184 ( 0.2623) = 0.356 J يسخر من الجروح من ال يعرف األلم. )5(
A point charge q 1 is held stationary at the origin. A second charge q 2 is placed at point a, and the electric potential energy of the pair of charges is +5.4 10 8 J. When the second charge is moved to point b, the electric force on the charge does 1.9 10 8 J of work. What is the electric potential energy of the pair of charges when the second charge is at point b? w a b = U a U b U a = +5.4 10 8 w a b = 1.9 10 8 = U a U b U b = U a w a b = 1.9 10 8 ( 5.4 10 8 ) = 7.3 10 8 J ال تشغل بالك إال بما تستطيع, افعل ما تستطيع اآلن, وعندما تستطيع أكثر حينها افعل المزيد. )6(
(a) How much work would it take to push two protons very slowly from a separation of 2.00 10 10 m (a typical atomic distance) to 3.00 10 15 m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation? (a) U = 1 qq ο 4πε ο r W = ΔU = U 2 U 1 (b) = 1 4πε ο ( e2 r 2 e2 r 1 ) = (9.00 10 9 )(1.60 10 19 ) 2 1 ( 3.00 10 15 1 2.00 10 15) = 7.68 10 14 J ΔU = K 1 + K 2 = 2K = 2 ( 1 2 mv2 ) = mv 2 v = ΔU m = 7.68 10 14 1.67 10 27 = 6.78 106 m/s من الطبيعي أال تكسب كل مرة. )7(
A small metal sphere, carrying a net charge of q 1 = 2.80 μc, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q 2 = 7.80 μc and mass 1.50 g, is projected toward q 1. When the two spheres are 0.800 m apart, q 2 is moving toward q 1 with speed 22.0 m/s (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of q 2 when the spheres are 0.400 m apart? (b) How close does q 2 get to q 1? W other = 0, U = 1 q 1 q 2 4πε ο r K a = 1 2 mv a 2 = 1 2 (1.50) 10 3 )(22.0) 2 = 0.3630 J U a = 1 q 1 q 2 = (8.988 10 9 ) ( 2.80 10 6 )( 7.80 10 6 ) = 0.2454 J 4πε ο r a 0.800 K b = 1 2 mv b 2, U b = 1 4πε ο q 1 q 2 r b U b = (8.988 10 9 ) ( 2.80 10 6 )( 7.80 10 6 ) 0.400 K b = K a + (U a + U b ) 1 2 mv b 2 = 0.3630 + (0.2454 0.4907) = 0.1177 J v b = 2(0.1177) = 12.5 m/s 1.50 10 3 )8( = 0.4907 J ال تنافس إال على القمة.
Three equal 1.20 μc point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) U = 3kq2 0.500 = 3k(1.2 10 6 ) 2 0.500 = 0.078 J إذا أسديت جميال إلى إنسان فاحذر أن تذكره وإن أسدى إنسان إليك جميال فاحذر أن تنساه. )عبدهللا بن المقفع(. )9(
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square? W center side = U center U side = 4 1 4πε ο q α q e r 1 (2 1 4πε ο q α q e r 2 + 2 1 4πε ο q α q e r 3 ) Substituting q e = e and q α = 2e W center side = 4 e 2 1 4πε ο [ 2 r 1 ( 1 r 2 + 1 r 3 )] W = 4(1.6 10 19 ) 2 [ 2 50 ( 1 5.00 + 1 125 )] = 6.08 10 21 J لعله من عجائب الحياة, إنك إذا رفضت كل ما هو دون الق مة, فإنك ستحصل عليه. )11(
A small particle has charge 5.00 μc and mass 2.00 10 4 kg. It moves from point A, where the electric potential is V A = +200 V, to point B, where the electric potential is V B = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain. K A + U A = K B + U B U = qv K A + qv A = K B + qv B K B = K A + q(v A V B ) = 0.00250 + ( 5.00 10 6 )(200 800) = 0.00550 J v B = 2K B m = 7.42 m/s ما يم كن تخي له يم كن تحقيقه, وما يمكن تحقيقه لن نعدم طري قا للوصول إليه. )11(
Two stationary point charges +3.00 nc and +2.00 nc are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00 nc charge? K a + U a = K b + U b keq 1 r 1a keq 2 r 2a = keq 1 r 1b keq 2 r 2b + 1 2 m ev b 2 E a = K a + U a = k( 1.60 10 19 ) [ (3.00 10 9 ) 0.250 + (2.00 10 9 ) ] = 2.88 10 17 J 0.250 E b = K b + U b = k( 1.60 10 19 ) [ (3.00 10 9 ) 0.100 + (2.00 10 9 ) ] + 1 0.400 2 m 2 ev b = 5.04 10 17 J + 1 2 m ev b 2 Setting E a = E b 2 v b = 9.11 10 31 (5.04 10 17 2.88 10 17 ) النجاح عادة وكذلك 10 6 m/s 6.89 = يكون الفشل. )12(
Two point charges q 1 = +2.40 nc and q 2 = 6.50 nc are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q 1 and 0.060 m from q 2 (Fig. E23.19). Take the electric potential to be zero at infinity. Find (a) the potential at point A; (b) the potential at point B; (c) the work done by the electric field on a charge of 2.50 nc that travels from point B to point A. (a) V A = 1 4πε ο ( q 1 r A1 + q 2 r A2 ) V A = (8.988 10 9 2.40 10 9 ) ( + 0.050 = 737 V (b) V B = 1 ( q 1 + q 2 ) 4πε ο r B1 r B2 V B = (8.988 10 9 2.40 10 9 ) ( + 0.080 (C) = 704 V W B A = q (V B V A ) 6.50 10 9 ) 0.050 6.50 10 9 ) 0.060 ر ب سكوت أبلغ 10 8 J 8.2 = )) ( 737 )( 704 9 10 (2.50 = من كالم. )13(
At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 12.0 V c, respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge? E = k q kq and V = r2 r (a) V > 0 SO q > 0 V E = kq/r k q /r 2 = (kq r ) (r2 kq ) = r r = 4.98 = 0.415 m 12.0 (b) q = rv K = (0.415)(4.98) 8.99 10 9 = 2.30 10 10 C (c) q > 0 So the electric field is directed away from the charge. )14( ت ج ن ب الخطأ أسهل من طلب العفو.
A total electric charge of 3.50 nc is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm; (b) 24.0 cm; (c) 12.0 cm. (a) V = kq r = k(3.50 10 9 ) 0.480 = 65.6 V (b) V = k(3.50 10 9 ) 0.240 = 131 V (c) This is inside the sphere. The potential has the same value as at the surface, 131 V. الفاشلون يعتقدون أن النجاح مجرد عملية حظ. )15(
A very long wire carries a uniform linear charge density λ. Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 cm from the wire and the other probe is 1.00 cm farther from the wire, the meter reads 575 V. (a) What is λ? (b) If you now place one probe at 3.50 cm from the wire and the other probe 1.00 cm farther away, will the voltmeter read 575 V? If not, will it read more or less than 575 V? Why? (c) If you place both probes 3.50 cm from the wire but 17.0 cm from each other, what will the voltmeter read? (a) V = λ ln(r 2πε b r a ) ο λ = ( V)2πε ο ln(r b r a ) = 575 (18 10 9 )ln ( 3.5 2.50 ) = 9.49 10 8 C/m (b) The meter will read less than 575 V because the electric field is weaker over this 1.00-cm distance than it was over the 1.00-cm distance in part (a). (c) The potential difference is zero because both probes are at the same distance from the wire, and hence at the same potential. ال يصمد مع العزيمة شيء. )16(
A very long insulating cylindrical shell of radius 6.00 cm carries charge of linear density 8.50 μc m spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 cm above the surface, and (b) the surface and a point 1.00 cm from the central axis of the cylinder? (a) V = λ ln(r 2πε b r a ) ο = (8.50 10 6 )(2 9.00 10 9 )ln ( 10.0 6.00 ) = 7.82 10 4 V = 78.200 V = 78.2 kv (b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero. تعتقد بعض النساء أن الزواا هو الفرصة الوحيدة لالنتقام من الرجل. )17(
Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate has magnitude 47.0 nc m 2, what is the magnitude of E in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference? (a) E = σ ε ο = 47.0 10 9 ε ο = 5310 N/C (b) V = Ed = (5310)(0.0220) = 117 V (c) The electric field stays the same if the separation of the plates doubles. The potential difference between the plates doubles. ما يقلق أغلبنا نظرة أناس قد ال يكادون يهتمون بنا على اإلطالق. )18(
(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kv? (b) What is the potential of the sphere s surface relative to infinity? (a) E = Q 4πε ο R Q = 4πε ο RV (b) = (0.125)(1500)(9.00 10 9 ) = 2.08 10 8 C = 20.8 nc Since the potential is constant inside the sphere, its value at the surface must be the same as at the center, 1.5 kv. أن تضيء شمعة صغيرة خير لك من أن تنفق عمرك تلعن الظالم. )19(
A particle with charge +7.60 nc is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right After it has moved 8.00 cm, the additional force has done 6.50 10 5 J of work and the particle has 4.35 10 5 J of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field? (a) W tot = K = K b K a = 4.35 10 5 J W FE = W tot W F = 4.35 10 5 6.5 10 5 = 2.15 10 5 J (b) W a b = q(v a V b ) V a V b = W a b q (c) = 2.15 10 5 7.60 10 9 = 2.83 103 V W a b = F E d = qed E = W a b qd = V a V b d = 2.83 103 0.0800 = 3.54 10 4 V/m أغلب المشاكل مؤقتة ونتائجها محدودة. )21(
A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 cm apart (Fig. P23.62). The plates are insulating and have uniform surface charge densities +σ and σ. The charge on the sphere is q = 8.90 10 6 C. What potential difference between the plates will cause the thread to assume an angle of 30.0 o with the vertical? T cos θ = mg T sin θ = F e F e = mg sin θ / cos θ = (1.50 10 3 )(9.80) tan 30 = 0.0085 N F e = Eq = Vq d V = Fd q = (0.0085)(0.0500) 8.90 10 6 = 47.8 V أصبح غالبيتنا يتطلع إلى غيره ليحدد له ماذا يفعل, وماذا ينبغي عليه أن يأكل وأن يفكر فيه. )21(
An insulating spherical shell with inner radius 25.0 cm and outer radius 60.0 cm carries a charge of +150.0 μc uniformly distributed over its outer surface (see Exercise 23.41). Point a is at the center of the shell, point b is on the inner surface, and point c is on the outer surface. (a) What will a voltmeter read if it is connected between the following points: (i) a and b; (ii) b and c; (iii) c and infinity; (iv) a and c? (b) Which is at higher potential: (i) a or b; (ii) b or c; (iii) a or c? (c) Which, if any, of the answers would change sign if the charges were 150 μc? (a) Points a, b and c are all at the same potential, so V a V b = V b V c = V a V c = 0 V c V = kq R = (8.99 109 )(150 10 6 ) = 2.25 10 6 V 0.60 (b) They are all at the same potential. (c) Only V c V would change; it would be 2.25 10 6 V. طريق النجاح يمر فوق بعض العقبات. )22(
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius three times that of sphere B. Let Q A and Q B be the charges on the two spheres, and let E A and E B be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio Q B Q A and (b) the ratio E B E A? (a) V = 1 4πε ο q R V A = Q A 4πε ο R A and V B = Q B 4πε ο R B V A = V B Q A 4πε ο R A = Q B 4πε ο R B Q B Q A = R B R A, R A = 3R B, Q B Q A = 1 3 (b) E = 1 q 4πε ο R 2 E A = ( Q A 4πε ο R 2 ) and E B = ( Q B A 4πε ο R 2 ) B E B = ( Q B E A 4πε ο R 2 ) ( 4πε οr 2 A ) B Q A = Q B ( R 2 A ) Q A R B = ( 1 3 ) (3)2 = 3 قد تنسى من شاركك الضحك, لكنك لن تنسى من شاركك البكاء. )23(