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10TH CLASSES PHYSICS DAILY PLAN Ex: SUBJECT: ELECTRIC FIELD & POTENTIAL GOALS: DURATION: IN PRACTICE: PRESENTATION: The Electric Field : An electric field at a point in space can be defind in terms of the negative electric force acting on a test charge q0 placed at that point. E q0 q a) Find the electrostatic force acting on q. 370 b) If q=-1/2 C find the direction & the magnitude of E. q W=12N Electric Field Lines: A more useful way to describe an electric field is with electric field lines, also called lines of force. * Each line shows the path that would be followed by a free positive test charge released in the field. ( q>>q0 ) q0 E q0 q Since F k . 2 d F E q0 E k. E i) E is tangent to the electric field line at each point. ii) The field is strong where the lines are close together, it is weaker where the lines are far apart. q.q 0 d2 q0 E k. q d2 * The electric field is a vector quantity having the SI unit of (N/C) * The direction of at a point is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point. * * F=E.q0 +q F -q F E Ex: q1=q a K a q3=-3q a q2=-3q If the electric field produced by q1 at point K is defined as E, then find the net electric field at K in terms of E. * The rules for drawing electric field lines for any charge distribution are: i) The lines must begin on positive charges and terminate on negative charges or at infinity in the case of an excess of charge. ii) The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitute of the charge. iii) No two field lines can cross. A +2q x B -q Fe x qEx SI→ W is Joule * W K qEx 0 qEx 1 1 2 2 mVB mV A qEx 2 2 W E K Motion of Charged Particles in a Uniform Electric Field: E & (J) (0,0) V V0=0 E (x,y) V * V X V0 cons tan at X F=q.E=m.a v y at a * q.E m x v0 .t y * If E is uniform (that is constant in magnitude &direction) - the acceleration of the motion is constant. - if the charge is positive the acceleration will be in the direction of the electric field. - if the charge is negative the acceleration will be in the opposite of the electric field. * V V0 at V V0 2ax 2 1 2 1 eE 2 at t 2 2 m * The trajectory is parabola. * After the electron leaves the region of uniform electric field it continues to move in a straight line with a speed v > v0 * Notice that we have neglected the gravitational force on the electron. (in E=104N/C the ratio of “elec.force/grav.force” is 10 14N for an electron and 1011N for a proton) 1 x x 0 V0 t at 2 2 2 eE 2 .t m NOTE: The field inside a charged conductor is always zero. Taking x0=0 & V0=0 gives; qE 2 1 x at 2 x t 2 2m qE t ; V at V m (1) (2) 2qE V 2 2ax V 2 x m * The kinetic energy of the charge after it has moved a distance x is; K 1 1 2qE mV 2 m x K q x 2 2 m * Also from Work-Energy Theorem; (2) (4) * The entire charge of the ball is transferred to the outer surface of the sphere. E q r E0 k q r2 d (Electric field of a charged conducting sphere as a function at the distance from its center.) HOMEWORK: MULTIMEDIA: DEMONSTRATION: EXPERIMENT: TEACHER: DIRECTOR: