5- In free space, a semi-circle ring with radius 'a' is carrying a uniform line charge distribut

First, we need to determine the electric field intensity at the origin due to a small element of

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5- In free space, a semi-circle ring with radius 'a' is carrying a uniform line charge distribution \(\rho_l = \rho_o \) [C/m]. The semi-circle ring is located in the x-y plane with its center being the origin as shown in the side figure. The electric field intensity vector in [V/m] at the origin is: a) \(\frac{\rho_o}{4\pi\epsilon_o a} (\hat{a}_x + \hat{a}_y)\) b) \(\frac{\rho_o}{2\pi\epsilon_o a^2} \hat{a}_y\) c) \(\frac{\rho_o}{2\pi\epsilon_o a} \hat{a}_y\) d) \(\frac{-\rho_o}{2\pi\epsilon_o a} \hat{a}_y\) e) \(\frac{\rho_o}{4\pi\epsilon_o a} \hat{a}_y\) f) None of the above

          5- In free space, a semi-circle ring with radius 'a' is carrying a uniform line charge distribution \(\rho_l = \rho_o \) [C/m]. The semi-circle ring is located in the x-y plane with its center being the origin as shown in the side figure. The electric field intensity vector in [V/m] at the origin is:
a) \(\frac{\rho_o}{4\pi\epsilon_o a} (\hat{a}_x + \hat{a}_y)\)
b) \(\frac{\rho_o}{2\pi\epsilon_o a^2} \hat{a}_y\)
c) \(\frac{\rho_o}{2\pi\epsilon_o a} \hat{a}_y\)
d) \(\frac{-\rho_o}{2\pi\epsilon_o a} \hat{a}_y\)
e) \(\frac{\rho_o}{4\pi\epsilon_o a} \hat{a}_y\)
f) None of the above

5- In free space, a semi-circle ring with radius 'a' is carrying a uniform line charge distribution = [C/m]. The semi-circle ring is located in the x-y plane with its center being the origin as shown in the side figure. The electric field intensity vector in [V/m] at the origin is:
a) ()/(4π a) (âx + ây)
b) ()/(2π a^2)ây
c) ()/(2π a)ây
d) (-)/(2π a)ây
e) ()/(4π a)ây
f) None of the above

Added by Michael M.

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