6.2 Electric field due to a ring of charge

  Electrostatics

Consider a conducting ring of radius $R$ having a total charge $q$ which is uniformly distributed over its circumference.

The electric field due to an infinitesimally small section $dl$ having charge $dq$ at any point $P$ which is at a distance $x$ from the center of the ring is given by, $$d\overrightarrow E = \frac{{kdq}}{{{r^2}}}$$

Resolving electric field into a horizontal and vertical component.
$$d\overrightarrow E = d{\overrightarrow E _x} + d{\overrightarrow E _y}$$
Integrating both sides, $$\int {d\overrightarrow E } = \int {d{{\overrightarrow E }_x}} + \int {d{{\overrightarrow E }_y}} $$

If we consider top and bottom segments of the ring, we see that the contributions of the field at point $P$ from these segments have same $x-$ component but opposite $y-$ components. Hence, the total $y-$ component of field due to this pair of segments is zero.
$${\int {d{{\overrightarrow E }_y}} = 0}$$ So, $$\int {d\overrightarrow E } = \int {d{{\overrightarrow E }_x}} $$ As $d{E_x} = dE\cos \theta $, $$\int {dE} = \int {\frac{{kdq}}{{{r^2}}}\cos \theta } $$ Also, $\cos \theta = \frac{x}{r} = \frac{x}{{\sqrt {{x^2} + {R^2}} }}$

$$\int {dE} = \int {\frac{{kdqx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}} $$ or $$\int {dE} = \frac{{kx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}\int {dq} $$ or $$E = {E_x} = \frac{{kqx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}} = \frac{{qx}}{{4\pi {\varepsilon _0}{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}}$$ and $${E_y} = 0$$
So, the electric field is only along $x-$ axis.

Note:

• $E=0$ at $x=0$. Field is zero at the centre of the ring.
  
  
• At very large distance from the center of ring i.e., $x > > R$. So, $\left( {{x^2} + {R^2}} \right) \approx {x^2}$ Therefore, $$E = \frac{{kq}}{{{x^2}}}$$
  So, the ring behaves like a point charge to an observer far from the ring.
  
  
• $E$ will be maximum where $\frac{{dE}}{{dx}} = 0$. So, $$\frac{{dE}}{{dx}} = kq\left[ {\frac{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}} - \frac{3}{2}x\left( {2x} \right){{\left( {{x^2} + {R^2}} \right)}^{\frac{1}{2}}}}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}}} \right]$$$$kq{\left( {{x^2} + {R^2}} \right)^{\frac{1}{2}}}\left[ {\frac{{\left( {{x^2} + {R^2}} \right) - 3{x^2}}}{{{{\left( {{x^2} + {R^2}} \right)}^3}}}} \right] = 0$$$$x = \frac{R}{{\sqrt 2 }}$$ So, $${E_{\max }} = \frac{{kqx}}{{{{\left( {{x^2} + {R^2}} \right)}^{\frac{3}{2}}}}} = \frac{{kq\left( {\frac{R}{{\sqrt 2 }}} \right)}}{{{{\left( {\frac{{{R^2}}}{2} + {R^2}} \right)}^{\frac{3}{2}}}}} = \frac{{2kq}}{{{{\left( 3 \right)}^{\frac{3}{2}}}{R^2}}}$$