Electric field Intensity due to a uniformly charged ring

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Electric field Intensity due to a uniformly charged ring

 

Electric field Intensity due to a uniformly charged ring can be evaluated at two points, one at its Centre and other at a point on its axis.

Case i :- Electric field Intensity due to a uniformly charged ring at its Centre

Electric field Intensity due to a uniformly charged ring - Curio Physics

Here by symmetry we can say that electric field intensity at Centre due to every small segment on ring is cancelled by the electric field intensity at Centre due to the segment exactly opposite to it. The electric field intensity at Centre due to segment AB is cancelled by that due to segment CD. Thus net electric field intensity at the Centre of a uniformly charged ring is E0=0

 

Case ii :- Electric field Intensity due to a uniformly charged ring at a point on it’s axis

Here we’ll find the electric field intensity at a point P due to a uniformly charged ring which is situated at a distance x from the center “O” of the ring. For this we consider two diametrically opposite small sections of equal length dl (in blue color) on ring as shown in figure below.

Electric field Intensity due to a uniformly charged ring - Curio Physics

The charge on these elemental sections is given by \displaystyle dq=\frac{Q}{2\pi R}dl

Where Q = total charge on the ring

Due to the element dl, electric field strength dE at point P can be given by \displaystyle \overrightarrow{dE}=\frac{kdq}{\left( {{R}^{2}}+{{x}^{2}} \right)}

The component of this field intensity dE sinθ which is normal to the axis of ring will be cancelled out due to the ring’s diametrically opposite section’s field component dE’ sinθ.

The component of electric field intensity along the axis of ring dEcosθ due to all the sections will be added up. Hence total electric field intensity at point P due to the ring is given by :-

\displaystyle E=\int{dE\cos \theta =\int\limits_{0}^{2\pi R}{\frac{kdq}{\left( {{R}^{2}}+{{x}^{2}} \right)}}}\times \frac{x}{\sqrt{{{R}^{2}}+{{x}^{2}}}}

\displaystyle E=\int\limits_{0}^{2\pi R}{\frac{kQx}{2\pi R{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}}dl

\displaystyle \Rightarrow E=\frac{kQx}{2\pi R{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\int\limits_{0}^{2\pi R}{dl}

\displaystyle \Rightarrow E=\frac{kQx}{2\pi R{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\left[ 2\pi R-0 \right]

\displaystyle \therefore E=\frac{kQx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}

In vector form,

\displaystyle \overrightarrow{E}=\frac{kQx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\widehat{i}

 

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Comments (2) on “Electric field Intensity due to a uniformly charged ring”

  1. Amazing explanation with breif and timely helpful efficient notes.. Curio physics has done some magical work out here by helping those fighting with physics and especially I’ll like to give special thanks to Mr. Manoj Kumar Verma, the genius mind behind such a world class educational content .. All I can say is a big thank you to you sir, Mr. Manoj Kumar Verma

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