CHAPTER 21 ELECTRIC CHARGE AND ELECTRIC FIELD...

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1 CHAPTER 21 ELECTRIC CHARGE AND ELECTRIC FIELD TWO BASIC CONCEPTS: COULOMB’S LAW ELECTRIC FIELD

Transcript of CHAPTER 21 ELECTRIC CHARGE AND ELECTRIC FIELD...

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CHAPTER 21 

ELECTRIC CHARGE AND 

ELECTRIC FIELD 

TWO BASIC CONCEPTS: 

COULOMB’S LAW 

     

ELECTRIC FIELD 

     

 

   

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ATOMS – are neutral 

Fig 21.4 

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IONS – are charged 

 

 

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SOME MATERIALS –  

e’s move easily – conductors 

metals 

 

SOME MATERIALS –  

  e’s don’t move easily –  

                insulators 

                glass 

                wood 

     

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FORCES ON CHARGED OBJECTS 

LIKE CHARGES REPEL 

UNLIKE CHARGES ATTRACT 

 

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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. 

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Attractive Force 

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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. 

       

   

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ELECTRIC FIELD 

 

Electric force per unit charge. 

 

   

 

Or 

 

   

 

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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. 

 

 

   

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For two charges q1 and qtest 

 

 

 

So  

 

       

     

Therefore 

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Or 

 

     

 

 

   

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ALSO 

       

 

Where    

 

 

Therefore can write 

 

     

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And 

 

     

 

   

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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. 

 

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Insert fig 21.29 

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Example 21.2 (continued) 

 

Find the electric field 3.0 cm 

from an electric charge  

q1 = +25 nC. 

 

  r = 3 cm 

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What is E at distance r due to q1 

 

     

 

       

 

DIRECTION ? 

 

Away from positive q1 

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If we place a charge ‐75nC at a 

point 3 cm away from q1 what 

is the force on this charge? 

 

 

 

 

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The electric field at q2 is 

 

Therefore 

 

 

 

 

 

WE OBTAINED FORCE IN TWO 

WAYS 

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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). 

. A 

                            300 

                               

.                            . X 

Q2                         Q1 

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          (towards Q1) 

 

   

  

                 (away from Q2) 

 

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EA2         

          EA               

                        

 

A     EA1 

     300 

 

 

 

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NOW CALCULATE THE FORCE ON A 

CHARGE Q3 = 50µC PLACED AT POINT A. 

 

 

 

 

WHAT DIRECTION? 

   

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HAVE BEEN TALKING ABOUT POINT 

CHARGES. 

 

WHAT ABOUT CONTINUOUS CHARGES? 

 

 

   

 

 

   

 

 

 

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Consider a ring of uniform charge 

 

Fig 21.24 

 

BY SYMETRY Ey = 0 

 

 

  

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Where   

 

 

 

 

 

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Put in for   

 

 

 

 

 

Finally 

 

 

 

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THIS IS THE ELECTRIC FIELD AS A 

FUNCTION OF x FOR A RING OF CHARGE 

WITH RADIUS a. 

 

   

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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 

 

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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 

 

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TO FIND THE TOTAL FIELD INTEGRATE 

USE THE CHARGE DENSITY σ 

 

THEN 

 

 

 

CHARGE ON RING 

 

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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 

 

 

 

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LET RADIUS EXPAND TO A VALUE OF 

INFINITY 

 

 

 

 

 

 

 

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ANYWHERE NEAR AN INFINITE SHEET OF 

CHARGE THE ELECTRIC FIELD WILL BE