Question

$\sf \red{ Question }$
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).
‏‏‎ ‎
$\sf \red{ Note \: :− }$
• Answer it correctly.
• Please don't spam.
• Need full explanation.
• Don't give any irrelevant answer.
━━━━━━━━━━━━━━━━━━━━━━━━━
​

Answer 1

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

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

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

$\longrightarrow\displaystyle \sf E_x = \dfrac{k l}{( {r}^{2} + {l}^{2})^{ ^{ \frac{3}{2} } } } \int\limits_{0}^{Q} dq$

$\footnotesize\longrightarrow \displaystyle \underline{ \boxed{\sf E_{net}= E_x = \dfrac{k Ql}{( {r}^{2} + {l}^{2})^{ ^{ \frac{3}{2} } } } }}$

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:

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

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

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

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

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

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

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

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

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

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

$\footnotesize\longrightarrow\sf {r}^{2} - 2 {l}^{2} = 0$

$\footnotesize\longrightarrow\sf {r}^{2} = 2 {l}^{2}$

$\footnotesize\longrightarrow \gray{\underline{ \boxed{ \orange{\bold{ l = \pm \dfrac{ r}{ \sqrt{2} }}}} }}$

Maximum Electric Field at l = r/√2:

$\footnotesize\longrightarrow\:\sf E_{net}= \dfrac{kQl}{ (r^2 + { l}^{2} )^{ \frac{3}{2} }}$

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

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

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

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

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

Answer 2

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).

‏‏‎ ‎

To find the electric field strength at a point on the axis of a charged ring, we can use the principle of superposition and consider the contributions to the field from all the small segments of the ring. The magnitude of the electric field at a distance / from the center of the ring is given by:

E(/) = kq(z/sqrt(r^2 + z^2)^3)

where k is Coulomb's constant, q is the charge on the ring, z is the distance from the center of the ring to the point on the axis, and r is the radius of the ring.

When I >>> r, we can approximate the expression for E(/) as:

E(/) ≈ kqz/I^3

where I is the moment of inertia of the ring, defined as I = mr^2, where m is the mass of the ring.

To find the maximum magnitude of the electric field strength, we can take the derivative of E(/) with respect to z and set it to zero:

dE(/)/dz = 3kqz(r^2 + z^2)^(-5/2) = 0

This gives us z = r/sqrt(2), which is the distance on the axis where the field is maximum. Plugging this back into the expression for E(/), we get:

E(max) = kq/(r(2)^(3/2))

To plot the function E(/), we can use a graphing calculator or software. The plot should show that the field is zero at the center of the ring, increases as we move away from the center, reaches a maximum at a distance of r/sqrt(2), and then decreases as we move further away from the ring.

Overall, the expression for the electric field strength on the axis of a charged ring is a good example of how we can use principles of superposition and integration to calculate the electric field from complex charge distributions.

Hope This will help you! :)