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Tushar Watts Grade: 12
        

4516-1717_3899_elecro1.bmp


The figure shows three circular arcs centred on origin of coordinate system.On each arc, the uniformly distributed charge is given in terms of Q. The radii are given in terms of R.  What are the magnitude and direction   of the net electric field at the origin due to the arcs.

7 years ago

Answers : (1)

Pratham Ashish
17 Points
										

4517-1054_4788_3.bmpfirst we will find the expression for E at the centre due to an arc


resultant E due to the small charges element would be in the direction  . joining the midpoint of the arc to the centre


conider only the half arc , & take only the component parallel to the above mentioned directions


  E =∫ 2/4¶ε dq cos( 45-x)/ r^2                     dq = λdl = λr dx


 E =∫ 2/4¶ε λ cos( 45-x)/ r  dx


 E = λ/2¶ε r ∫cos( 45-x)  dx


 E =  λ/2¶ε r sin(45-x) ]


E =  λ/2¶ε r  *1/√2


 


charge density at the first arc ,  λ1 = +2Q/ ¶R


E1 = + 2Q/ ¶R 2¶ε R  *1/√2


  = +Q /√2 ε ¶^2 R^2 


 


λ2 = -4Q / ¶R


E2 = -Q /√2 ε ¶^2 R^2 


λ3 = 6Q/ ¶R


E3 = Q /√2 ε ¶^2 R^2


Resultant E = E1 +E2 +E3


                 =  Q /√2 ε ¶^2 R^2   to the direction joining the midpoint of the arc to the centre & towards the centre


 


 


 


 


 

7 years ago
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