Particle in Uniform Electric or Gravitational Field
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1 Particle in Uniform Electric or Gravitational Field particle charge mass electron q e = e m e = kg proton q p = +e m p = kg neutron q n = m n = kg Elementar charge: e = C. Electric field equation of motion: F = m a force law: F = q E acceleration: a = (q/m) E E q q n = F e p F p F n = q e Gravitational field equation of motion: F = m a g m m m p n e force law: F = m g acceleration: a = g F p F n F e 1/9/215 [tsl24 1/15]
2 Projectile Motion in Electric Field electrostatic force: F = F = ee equation of motion: acceleration: a = velocit: v (t) = v cos θ F = me a a = e m e E a v (t) = v sin θ at position: (t) = v [cos θ]t (t) = v [sin θ]t 1 2 at2 height: h = v2 2a sin2 θ range: R = v2 a sin(2θ) m e qe = e v θ R h E 1/9/215 [tsl26 2/15]
3 Cathode Ra Tube 1/9/215 [tsl419 3/15]
4 Particle Projected Perpendicular to Uniform Electric Field A charged particle (m = 3kg, q = 1µC) is launched at t = with initial speed v = 2m/s in an electric field of magnitude E = N/C as shown. m q E 1m 1m v (a) Find the position of the particle at t 1 = 3s. (b) B what angle does the velocit vector turn between t = and t 1 = 3s? 1/9/215 [tsl27 4/15]
5 Particles Accelerated b Uniform Electric Field A uniform electric field E = N/C eists in the bo. (a) A charged particle of mass m 1 = kg is released from rest at = 3cm, =. It eits the bo at = 3cm, = 6cm after a time t 1 = s. Find the charge q 1. (b) A second charged particle of mass m 2 = kg is projected from position =, = 3cm with initial speed v = m/s. It eits the bo at = 3.9cm, = 6cm. Find the charge q 2. 6 [cm] 4 q 2 E 2 v 2 q [cm] 1/9/215 [tsl28 5/15]
6 Action and Reaction due to Coulomb Interaction Two particles with masses m 1, m 2 and charges q 1, q 2 are released from rest a distance r apart. We consider the following four distinct configurations: (a) m 1 = 1kg, m 2 = 1kg, q 1 = 1C, q 2 = 1C (b) m 1 = 1kg, m 2 = 1kg, q 1 = 1C, q 2 = 2C (c) m 1 = 1kg, m 2 = 2kg, q 1 = 1C, q 2 = 1C (d) m 1 = 1kg, m 2 = 2kg, q 1 = 1C, q 2 = 2C m 1 r q 1 m 2 q 2 Anwer the following questions for each configuration: (1) Is the force eperienced b particle 1 smaller than or equal to or larger than the force eperienced b particle 2? (2) Is the acceleration of particle 1 smaller than or equal to or larger than the acceleration of particle 2? 1/9/215 [tsl29 6/15]
7 Particle in Uniform Electric and Gravitational Field (1) A proton, a neutron, and an electron are dropped from rest in a vertical gravitational field g and in a horizontal electric field E as shown. Both fields are uniform. p n e g E h (a) Which particle travels the shortest distance? (b) Which particle travels the longest distance? (c) Which particle travels the shortest time? (d) Which particle reaches the highest speed? 1/9/215 [tsl25 7/15]
8 Particle in Uniform Electric and Gravitational Field (2) A proton, a neutron, and an electron are dropped from rest in a vertical gravitational field g and in a horizontal electric field E as shown. Both fields are uniform. g E (a) Which particle travels the shortest distance? (b) Which particle travels in a straight line? (c) Which particle travels the shortest time? (d) Which particle reaches the highest speed? e n p h h 1/9/215 [tsl42 8/15]
9 Is the Faster also the Quicker? Charged particles 1 and 2 are released from rest in a uniform electric field. (a) Which particle moves faster when it hits the wall? (b) Which particle reaches the wall more quickl? d = 2m 1 m = 1kg 1 q = 2C 1 E = 4N/C d = 1m 2 m = 1kg 2 q = 1C 2 1/9/215 [tsl372 9/15]
10 Intermediate Eam I: Problem #3 (Spring 6) Consider a region of uniform electric field as shown. A charged particle is projected at time t = with initial velocit as shown. Ignore gravit. (a) Find the components a and a of the acceleration at time t =. (b) Find the components v and v of the velocit at time t =. (c) Find the components v and v of the velocit at time t = 1.2s. (d) Find the components and of the position at time t = 1.2s. m =3g q = 6mC v = 2m/s 1/9/215 [tsl35 1/15]
11 Intermediate Eam I: Problem #3 (Spring 6) Consider a region of uniform electric field as shown. A charged particle is projected at time t = with initial velocit as shown. Ignore gravit. (a) Find the components a and a of the acceleration at time t =. (b) Find the components v and v of the velocit at time t =. (c) Find the components v and v of the velocit at time t = 1.2s. (d) Find the components and of the position at time t = 1.2s. Solution: (a) a = q m E = C kg (5N/C) = 1m/s2, a =. m =3g q = 6mC v = 2m/s 1/9/215 [tsl35 1/15]
12 Intermediate Eam I: Problem #3 (Spring 6) Consider a region of uniform electric field as shown. A charged particle is projected at time t = with initial velocit as shown. Ignore gravit. (a) Find the components a and a of the acceleration at time t =. (b) Find the components v and v of the velocit at time t =. (c) Find the components v and v of the velocit at time t = 1.2s. (d) Find the components and of the position at time t = 1.2s. Solution: (a) a = q m E = C kg (5N/C) = 1m/s2, a =. (b) v =, v = v = 2m/s. m =3g q = 6mC v = 2m/s 1/9/215 [tsl35 1/15]
13 Intermediate Eam I: Problem #3 (Spring 6) Consider a region of uniform electric field as shown. A charged particle is projected at time t = with initial velocit as shown. Ignore gravit. (a) Find the components a and a of the acceleration at time t =. (b) Find the components v and v of the velocit at time t =. (c) Find the components v and v of the velocit at time t = 1.2s. (d) Find the components and of the position at time t = 1.2s. Solution: (a) a = q m E = C kg (5N/C) = 1m/s2, a =. (b) v =, v = v = 2m/s. (c) v = a t = (1m/s 2 )(1.2s) = 12m/s, v = v = 2m/s. m =3g q = 6mC v = 2m/s 1/9/215 [tsl35 1/15]
14 Intermediate Eam I: Problem #3 (Spring 6) Consider a region of uniform electric field as shown. A charged particle is projected at time t = with initial velocit as shown. Ignore gravit. (a) Find the components a and a of the acceleration at time t =. (b) Find the components v and v of the velocit at time t =. (c) Find the components v and v of the velocit at time t = 1.2s. (d) Find the components and of the position at time t = 1.2s. Solution: (a) a = q m E = C kg (5N/C) = 1m/s2, a =. (b) v =, v = v = 2m/s. (c) v = a t = (1m/s 2 )(1.2s) = 12m/s, v = v = 2m/s. m =3g q = 6mC v = 2m/s (d) = 1 2 a t 2 =.5(1m/s 2 )(1.2s) 2 = 7.2m, = v t = (2m/s)(1.2s) = 2.4m. 1/9/215 [tsl35 1/15]
15 Unit Eam I: Problem #3 (Spring 7) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2kg and charge q = 1C are projected at time t = with initial velocities as shown. Both particles will hit the screen eventuall. Ignore gravit. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height 2 does the particle in region (2) hit the screen? v = 2m/s (1) (2) screen v = 2m/s screen 8m 8m 1/9/215 [tsl361 11/15]
16 Unit Eam I: Problem #3 (Spring 7) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2kg and charge q = 1C are projected at time t = with initial velocities as shown. Both particles will hit the screen eventuall. Ignore gravit. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height 2 does the particle in region (2) hit the screen? Solution: (a) 1 = 1 2 at2 1 with a = q m E = 2.5m/s2, 1 = 8m t 1 = 2.53s. v = 2m/s (1) (2) screen v = 2m/s screen 8m 8m 1/9/215 [tsl361 11/15]
17 Unit Eam I: Problem #3 (Spring 7) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2kg and charge q = 1C are projected at time t = with initial velocities as shown. Both particles will hit the screen eventuall. Ignore gravit. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height 2 does the particle in region (2) hit the screen? Solution: (a) 1 = 1 2 at2 1 with a = q m E = 2.5m/s2, 1 = 8m t 1 = 2.53s. (b) 1 = v t 1 = 5.6m. v = 2m/s (1) (2) screen v = 2m/s screen 8m 8m 1/9/215 [tsl361 11/15]
18 Unit Eam I: Problem #3 (Spring 7) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2kg and charge q = 1C are projected at time t = with initial velocities as shown. Both particles will hit the screen eventuall. Ignore gravit. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height 2 does the particle in region (2) hit the screen? Solution: (a) 1 = 1 2 at2 1 with a = q m E = 2.5m/s2, 1 = 8m t 1 = 2.53s. (b) 1 = v t 1 = 5.6m. (c) 2 = v t 2 t 2 = 8m 2m/s = 4s. v = 2m/s (1) (2) 8m screen v = 2m/s 8m screen 1/9/215 [tsl361 11/15]
19 Unit Eam I: Problem #3 (Spring 7) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2kg and charge q = 1C are projected at time t = with initial velocities as shown. Both particles will hit the screen eventuall. Ignore gravit. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height 2 does the particle in region (2) hit the screen? Solution: (a) 1 = 1 2 at2 1 with a = q m E = 2.5m/s2, 1 = 8m t 1 = 2.53s. (b) 1 = v t 1 = 5.6m. (c) 2 = v t 2 (d) 2 = 1 2 at2 2 = 2m. t 2 = 8m 2m/s = 4s. v = 2m/s (1) (2) 8m screen v = 2m/s 8m screen 1/9/215 [tsl361 11/15]
20 Electric Dipole Field L E E + q +q p E =» kq ( L/2) 2 kq ( + L/2) 2 ( + L/2) 2 = kq ( L/2) 2 ( L/2) 2 ( + L/2) 2 2kqL 3 = 2kp 3 (for L) = 2kqL ( 2 L 2 /4) 2 Electric dipole moment: p = q L Note the more rapid deca of the electric field with distance from an electric dipole ( r 3 ) than from an electric point charge ( r 2 ). The dipolar field is not radial. 1/9/215 [tsl23 12/15]
21 Water Molecule 1/9/215 [tsl51 13/15]
22 Force and Torque on Electric Dipole The net force on an electric dipole in a uniform electric field vanishes. However, this dipole eperiences a torque τ = p L that tends to align the vector p with the vector E. Now consider an electric dipole that is alread aligned (locall) with a nonuniform electric field. This dipole eperiences a net force that is alwas in the direction where the field has the steepest increase. 1/9/215 [tsl328 14/15]
23 Electric Quadrupole Field L = L E E + +q 2q +q p p 2 1 E = = kq 2 kq ( L) 2 + kq ( + L) 2 + k( 2q) kqL2 4 2L + 3L2 2 + = 3kQ 4 (for L) = kq 2 1 " 1 L 2L + 3L2 2 q « L «# 2 2 «2 +q Electric quadrupole moment: Q = 2qL 2 Different quadrupole configuration: L L +q q. 1/9/215 [tsl327 15/15]
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