Respuesta :
Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
      [tex]\theta = \frac{S}{r}[/tex]
where, Â S = length of the rod
       r = radius
Putting the given values into the above formula as follows. Â Â Â
        [tex]\theta = \frac{S}{r}[/tex]
               = [tex]\frac{4}{2}[/tex]
               = [tex]2 radians (\frac{180^{o}}{\pi})[/tex]
               = [tex]114.64^{o}[/tex]
Now, we will calculate the charge density as follows.
         [tex]\lambda = \frac{Q}{L}[/tex]
              = [tex]\frac{18 \times 10^{-9} C}{4 m}[/tex]
              = [tex]4.5 \times 10^{-9} C/m[/tex]
Now, at the center of arc we will calculate the electric field as follows.
         E = [tex]\frac{2k \lambda Sin (\frac{\theta}{2})}{r}[/tex]
           = [tex]\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}[/tex]
           = 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
