The electric field along the [tex]z[/tex] axis is [tex]E\left( z \right) = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}[/tex]
The radius of the ring is: [tex]a[/tex]
The charge on the ring is: [tex]q[/tex]
The coulomb's constant is: [tex]k = \frac{1}{{4\pi {\varepsilon _0}}}[/tex]
The expression of the electric field is: [tex]E = \frac{{kq}}{{{r^2}}}[/tex] .......(i)
Here,
[tex]r[/tex] ⇒ distance = [tex]\sqrt {{z^2} + {a^2}}[/tex]
Now, putting [tex]r[/tex] in equation (i), we get,
[tex]\begin{array}{c}\\E = \frac{{kq}}{{{{\left( {\sqrt {{z^2} + {a^2}} } \right)}^2}}}\\\\ = \frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}\\\end{array}[/tex]
The electric field along the z-direction is
[tex]dE\left( z \right) = dE\cos \theta[/tex] ........(ii)
here, [tex]\cos \theta = \frac{z}{{\sqrt {{z^2} + {a^2}} }}[/tex]
Integrating equation (ii), we get,
[tex]\begin{array}{c}\\\int {d{E_z}} = \int {dE\cos \theta } \\\\E\left( z \right) = E\cos \theta \\\end{array}[/tex].......(iii)
Now substituting the values in equation (iii), we get,
[tex]\begin{array}{c}\\E\left( z \right) = \left( {\frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}} \right)\left( {\frac{z}{{\left( {\sqrt {{z^2} + {a^2}} } \right)}}} \right)\\\\ = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\\\end{array}[/tex]
Therefore, the electric field along the [tex]z[/tex] axis is [tex]E\left( z \right) = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}[/tex]
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Consider a uniformly charged ring in the [tex]xy[/tex] plane, centered at the origin. The ring has radius [tex]a[/tex] and positive charge [tex]q[/tex] distributed evenly along its circumference. What is the magnitude of the electric field along the positive [tex]z[/tex] axis? Use k in your answer, where [tex]k = \frac{1}{4 \pi \epsilon_0}[/tex]
