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``` 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.```
10 years ago

``` first 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

```
10 years ago
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