tex]• Answer it correctly.• Please don't spam. • Need full explanation. • Don't ...

Questions

August 2021 1 26 Report

[tex]\sf \red{ Question }[/tex]
A thin circular wire ring of radius r carries a charge q. Find the magnitude of electric field strength on the axis of the ring as a function of distance / from its centre. Investigate the obtained function at I>>>r. Find the maximum magnitude of field strength and the corresponding distance I. Also draw the approximate plot of the function E(I).
‏‏‎ ‎
[tex]\sf \red{ Note \: :− }[/tex]
• Answer it correctly.
• Please don't spam.
• Need full explanation.
• Don't give any irrelevant answer.
━━━━━━━━━━━━━━━━━━━━━━━━━

Answers & Comments

MystifiedBoy

Verified answer

Answer:

Electric field due to uniform charged ring on the axis :

Let,

• P = point under consideration

• r = radius of the ring

• dq = point charge

• l = Distance on P (on axis) from the centre of the ring

• R = distance between point charge 'dq' and Point under consideration 'P'

Consider an element of length 'dx' and point charge 'dq'.

Electric field at point 'P' due to point charge 'dq':

$\footnotesize\longrightarrow\:\sf dE = \dfrac{k (dq)}{R^2} \\$

By Pythagoras theorem :

$\bullet \rm \: R = \sqrt{ {r}^{2} + {l}^{2} }$

So,

$\footnotesize\longrightarrow\:\sf dE = \dfrac{k (dq)}{\sqrt{( {r}^{2} + {l}^{2} } )^{2} } \\$

$\footnotesize\longrightarrow\:\sf dE = \dfrac{k (dq)}{ {r}^{2} + {l}^{2} } \\$

• Total electric field at point P :

$\footnotesize\longrightarrow\:\sf \overrightarrow{E_p }= \overrightarrow{E_x} + \overrightarrow{E_y} \\$

• Electric Field in x direction :

$\footnotesize\displaystyle\longrightarrow\:\sf E_x = \int\limits dE_x \\$

$\footnotesize\displaystyle\longrightarrow\:\sf E_x = \int\limits dE \cos( \theta) \\$

$\footnotesize \longrightarrow\displaystyle \sf E_x = \int\limits \dfrac{k (dq)}{( {r}^{2} + {l}^{2}) } \cdot \cos( \theta) \\$

Cos(θ) = Base/Hypotenuse = l/(r² + l²)½

$\footnotesize \longrightarrow\displaystyle \sf E_x = \int\limits \dfrac{k (dq)}{( {r}^{2} + {l}^{2}) } \cdot \dfrac{l}{( {r}^{2} + {l}^{2})^{ \frac{1}{2} } } \\$

$\footnotesize\longrightarrow\displaystyle \sf E_x = \int\limits \dfrac{k (dq)l}{( {r}^{2} + {l}^{2})^{ ^{ \frac{3}{2} } } } \\$

$><img src=$

Electric field in y direction will be 0 because every element on the ring there is opposite element due to which perpendicular component of the electric field will cancel out each other.

• If l >>> r then,

$\footnotesize \longrightarrow\displaystyle \sf {r}^{2} + {l}^{2} \approx {l}^{2} \\$

Therefore, net electric field will be :

$\footnotesize\longrightarrow \displaystyle \underline{ \boxed{\sf E_{net}= \dfrac{k Ql}{ {l}^{3} } \: or \: E_{net}= \dfrac{k Q}{ {l}^{2} }}} \\$

• Value of 'l' such that Electric Field is maximum:

$\footnotesize\longrightarrow\sf E = Function \: of \: (l) \\$

$\footnotesize\longrightarrow\sf \dfrac{dE}{dl}= 0 \\$

$\footnotesize \longrightarrow\sf \dfrac{d}{dl}\Bigg( \dfrac{kQl}{(r^2 + l^2)^{\frac{3}{2}}}\Bigg)= 0$

$\footnotesize\longrightarrow\sf kQ \dfrac{d}{dl}\Bigg( \dfrac{l}{(r^2 + l^2)^{\frac{3}{2}}}\Bigg)= 0 \\$

$\dag\: \underline{\boxed{ \sf \dfrac{dy}{dx} = \dfrac{d}{dx}\bigg(\dfrac{u}{v }\bigg) = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx} }{ {v}^{2} }}}$

$\longrightarrow\sf kQ \Bigg( \dfrac{(r^2 + l^2)^{\frac{3}{2} } \dfrac{d}{dl}(l) - (l) \dfrac{d}{dl}(r^2 + l^2)^{\frac{3}{2} } }{(r^2 + l^2)^{\frac{3}{2} }.(r^2 + l^2)^{\frac{3}{2}}}\Bigg)= 0 \\$

$\footnotesize\longrightarrow\sf kQ \Bigg( \dfrac{(r^2 + l^2)^{\frac{3}{2} }(1) - (l) . \frac{3}{2} (r^2 + l^2)^{\frac{1}{2} } (2l) }{(r^2 + l^2)^3 }\Bigg)= 0 \\$

$><img src=$

$><img src=$

$\footnotesize\longrightarrow\:\sf E_{max_{ \bigg(at \: l = \frac{r}{ \sqrt{2} } \bigg )} }= \dfrac{kQ \bigg(\dfrac{r}{ \sqrt{2} } \bigg) }{ \bigg \{r^2 + \dfrac{r^2}{2}\bigg \}^ {\frac{3}{2}}}$

$\footnotesize\longrightarrow\:\sf E_{max_{ \bigg(at \: l = \frac{r}{ \sqrt{2} } \bigg )} }= \dfrac{kQ \bigg(\dfrac{r}{ \sqrt{2} } \bigg) }{ \bigg \{ \dfrac{3r^2}{2}\bigg \}^ {\frac{3}{2}}}$

$\footnotesize\longrightarrow\:\sf E_{max_{ \bigg(at \: l = \frac{r}{ \sqrt{2} } \bigg )} }= \dfrac{ \bigg(\dfrac{kQr}{ \sqrt{2} } \bigg) }{ \bigg \{ \sqrt{\dfrac{3r^2}{2}}\bigg \}^ {3}} \\$

I've skipped the basic calculation as the word limit is extended. It's easy you can do it by yourself.

$\footnotesize\longrightarrow\: \underline{ \boxed{ \red{\bf E_{max_{ \bigg(at \: l = \frac{r}{ \sqrt{2} } \bigg )} }= \dfrac{2kQ}{3\sqrt{3} r^2 } = \dfrac{ Q}{ 6 \sqrt{3} \pi \varepsilon_{0} {r}^{2} }} }} \\$