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Lecture 3-1 Dipole in uniform electric fields • No net force. The electrostatic forces on the constituent point charges are of the same magnitude but along opposite directions. So, there is no net force on the dipole and thus its center of mass should not accelerate. p E • Net torque! There is clearly a net torque acting on the dipole with respect to its center of mass, since the forces are not aligned. http://qbx6.ltu.edu/s_schneider/physlets/main/dipole_torque.shtml Lecture 3-2 Torque on the Dipole F The magnitude of the torque: Fx sin F (d x) sin Fd sin qEd sin pE sin x CM F The torque points into the screen. p E rF Lecture 3-3 Electric Field from Coulomb’s Law Bunch of Charges r + + + - qi+ - - 1 i P + - Continuous Charge Distribution P r dq + k qi E rˆ 2 i 4 0 i ri Summation over discrete charges http://www.falstad.com/vector3de/ 1 dq E rˆ d E 2 4 0 r dV dq dA dL (volume charge) (surface charge) (line charge) Integral over continuous charge distribution Lecture 3-4 Continuous Charge Distribution 1: Charged Line dx E Ex k L / 2 ( x x )2 p L/2 k xp L / 2 xp L / 2 du u2 u x xp L / 2 1 k u xp L / 2 k L 2 x p ( L / 2) 2 x p / L kQ / x p 2 At a point P on axis: p x xp L / 2 Q L Lecture 3-5 Again: Continuous Charge Distribution 1: Charged Line At a point P on perpendicular axis: E Ey L/2 k dx cos x y L/2 y k dx L / 2 ( x 2 y 2 ) 3/ 2 L/2 2 2 x y tan y 2 sec 2 k d ( y 2 tan 2 y 2 ) 3/ 2 k k cos d (2sin ) y y L / 2 and E 2k / y x 1 2 Lecture 3-6 Physics 241 –Warm-up quiz The rod is uniformly charged with a positive charge density . What is the direction of the electric field at a point P on a line perpendicular to the rod? Note that the line and the rod are in the same plane. a) b) c) d) e) to the right to the left up down lower right p Lecture 3-7 General location of P: Charged Line x2 dx x1 r2 Ex k sin k x2 x1 x dx 2 2 3/ 2 (x y ) At a point P off axis: x2 cos 2 cos 1 1 k 2 k 2 1/ 2 ( x y ) y y x1 k cos 2 cos1 y x2 Ey k x1 dx r 2 cos k x2 x1 y dx 2 2 3/ 2 (x y ) k 2 sec 2 k d sin 2 sin 1 2 3/ 2 y 1 (tan 1) y x y tan Lecture 3-8 Continuous Charge Distribution 2: Charged Ring At point P on axis of ring: ds Q 2 R Use symmetry! E Ex k ds x a 2 2 cos x k 2 ds 2 3/ 2 (x a ) x k 2 Q 2 3/ 2 (x a ) kQ / x 2 ( x a) (kQ / a 3 ) x ( x a) Lecture 3-9 Continuous Charge Distribution 3: Charged Disk Use the ring result x dE x k 2 dq 2 3/ 2 (x a ) At a point P on axis: x E E x dE x k 2 dq 2 3/ 2 (x a ) R x k 2 2 ada 2 3/ 2 0 (x a ) R a kx 2 da 2 2 3/ 2 0 (x a ) 2 1 / 2 R kx 2 ( x a ) 2 0 1 2 k 1 x 0 2 1 ( R / x ) R whole plane and E / 2 0 dq 2 a da Superposition of rings! 2 k x R 2 0 E 2 k R kQ (x 2 x 2 x <= Independent of x R) Lecture 3-10 Continuous Charge Distribution 4: Charged Sheets ( ) ( ) E=const in each region Superposition! Capacitor geometry Lecture 3-11 Physics 241 –Quiz 1A The left half of a rod is uniformly charged with a positive charge density , whereas the right half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular bisector and above the rod as shown? a) b) c) d) e) to the right to the left up down E is zeo. Lecture 3-12 Physics 241 –Quiz 1b The upper half of a ring is uniformly charged with a positive charge density , whereas the lower half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular axis through the center of the ring and to the left of the ring as shown? a) b) c) d) e) to the right to the left up down E is zeo. Lecture 3-13 Physics 241 –Quiz 1C The upper half of a rod is uniformly charged with a positive charge density , whereas the lower half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular bisector and to the right of the rod as shown? a) b) c) d) e) to the right to the left up down E is zeo.