CHAPTER 21 ELECTRIC CHARGE AND ELECTRIC FIELD TWO BASIC CONCEPTS: COULOMB’S LAW భ మ
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ELECTRIC FIELD 1 ATOMS – are neutral Fig 21.4 2 IONS – are charged 3 SOME MATERIALS – e’s move easily – conductors metals SOME MATERIALS – e’s don’t move easily – insulators glass wood 4 FORCES ON CHARGED OBJECTS LIKE CHARGES REPEL UNLIKE CHARGES ATTRACT 5 USE COULOMB’S LAW Example 21.2 Two point charges q1 = 25 nC and q2 = ‐75 nC are separated by a distance of 3.0 cm. Find the magnitude and direction of the electric force that q1 exerts on q2. 6 ଵ ଶ
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Attractive Force 7 Also Newton’s third law – if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. 8 ELECTRIC FIELD Electric force per unit charge. Or ೟೐ೞ೟
9 In direction that positive charge would move. For more than one charge the total electric field equals the vector sum of all electric fields due to each charge. 10 For two charges q1 and qtest So ೜భ ೜೟೐ೞ೟
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13 And బ
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14 ELECTRIC FIELD LINES 1. IN DIRECTION OF FIELD (POSITIVE TEST CHARGE MOVES ALONG LINE.) 2. NUMBER OF LINES PROPORTIONAL TO ELECTRIC FIELD. 3. ELECTRIC FIELD LINES START ON POSITIVE CHARGE AND END ON NEGATIVE CHARGE. 15 Insert fig 21.29 16 17 Example 21.2 (continued) Find the electric field 3.0 cm from an electric charge q1 = +25 nC. r = 3 cm 18 What is E at distance r due to q1 షవ
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DIRECTION ? Away from positive q1 19 If we place a charge ‐75nC at a point 3 cm away from q1 what is the force on this charge? 20 The electric field at q2 is ହ
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) WE OBTAINED FORCE IN TWO WAYS 21 Example‐ Field from two charges Q1 = ‐50µC at x=0.52m, y=0 Q2 = 50µC at x=0, y=0 and find E at point A (x=0, y=0.3m). y . A 0
30 . . X Q2 Q1 22 (towards Q1) షల
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(away from Q2) 23 EA2 EA A EA1 0 30
24 25 26 NOW CALCULATE THE FORCE ON A CHARGE Q3 = 50µC PLACED AT POINT A. ି଺
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WHAT DIRECTION? 27 HAVE BEEN TALKING ABOUT POINT CHARGES. WHAT ABOUT CONTINUOUS CHARGES? ௗொ
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28 Consider a ring of uniform charge Fig 21.24 BY SYMETRY Ey = 0 ௫
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32 THIS IS THE ELECTRIC FIELD AS A FUNCTION OF x FOR A RING OF CHARGE WITH RADIUS a. 33 START WITH THAT EQUATION AND FIND THE ELECTRIC FIELD ALONG THE AXIS OF A DISK WITH TOTAL CHARGE Q AND RADIUS OF R. INSERT FIG 21.26 ONCE AGAIN BY SYMETRY Ey = 0 34 Consider a small ring in the disk as shown. For the ring: Radius = r Thickness = dr Charge on ring = dQ WE CAN WRITE FOR THE CONTRIBUTION TO THE ELECTRIC FIELD AT P IN THE x DIRECTION 35 ௫
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TO FIND THE TOTAL FIELD INTEGRATE USE THE CHARGE DENSITY σ THEN ଶ
CHARGE ON RING 36 ௫
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37 ோ
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39 NOW CONSIDER AN INFINITE SHEET WITH CHARGE DENSITY σ WHAT IS THE ELECTRIC FIELD A DISTANCE x FROM THE SHEET? FOR A DISK OF RADIUS R ௫
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40 LET RADIUS EXPAND TO A VALUE OF INFINITY ௚௢௘௦
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41 ANYWHERE NEAR AN INFINITE SHEET OF CHARGE THE ELECTRIC FIELD WILL BE ଴
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