Question

qx
If E=
represents the electric field on
(a? + x2)3/2
the axis of the uniformly charged ring of radius 'a',
then value of E is maximum for what value of X on
the axis from the centre of the ring ?
a
(1) X= +
V2
(2) X = + a
(3) X = +
*
a
(4) X = +
22
2

please anyone tell the solution ​

Answer 1

Appropriate Question :

If $\sf E =\dfrac{kqx}{(a^2 + x^2)^{\frac{3}{2}}}$ represents the electric field on the axis of the uniformly charged ring of radius 'a' then value of E is maximum for what value of X on the axis from the centre of the ring ?

Solution :

$\maltese\:\underline{\textsf{\textbf{For maximum Electric Field :}}}$

$\longrightarrow\rm E = Function \: of \: (x)$

$\longrightarrow\rm \dfrac{dE}{dx}= 0$

$\longrightarrow\rm \dfrac{d}{dx}\Bigg( \dfrac{kqx}{(a^2 + x^2)^{\frac{3}{2}}}\Bigg)= 0$

$\longrightarrow\rm kq \dfrac{d}{dx}\Bigg( \dfrac{x}{(a^2 + x^2)^{\frac{3}{2}}}\Bigg)= 0$

Applying Division Rule :

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

Now,

$\longrightarrow\rm kq \Bigg( \dfrac{(a^2 + x^2)^{\frac{3}{2} } \dfrac{d}{dx}(x) - (x) \dfrac{d}{dx}(a^2 + x^2)^{\frac{3}{2} } }{(a^2 + x^2)^{\frac{3}{2} }.(a^2 + x^2)^{\frac{3}{2}}}\Bigg)= 0$

$\longrightarrow\rm kq \Bigg( \dfrac{(a^2 + x^2)^{\frac{3}{2} }(1) - (x) . \frac{3}{2} (a^2 + x^2)^{\frac{1}{2} } (2x) }{(a^2 + x^2)^3 }\Bigg)= 0$

$\longrightarrow\rm (a^2 + x^2)^{\frac{3}{2} }(1) - (x) . \frac{3}{2} (a^2 + x^2)^{\frac{1}{2} } (2x) = 0$

$\longrightarrow\rm (a^2 + x^2)^{\frac{3}{2} } - ( {3x}^{2} ) . (a^2 + x^2)^{\frac{1}{2} } = 0$

$\longrightarrow\rm (a^2 + x^2)^{\frac{3}{2} } . (a^2 + x^2)^{\frac{1}{2} } - 3 {x}^{2} = 0$

$\longrightarrow\rm (a^2 + x^2)^{\frac{1}{2} } . (a^2 + x^2)^{1 } - 3 {x}^{2} = 0$

$\longrightarrow\rm {a}^{2} + {x}^{2} - 3 {x}^{2} = 0$

$\longrightarrow\rm {a}^{2} - 2 {x}^{2} = 0$

$\longrightarrow\rm {a}^{2} = 2 {x}^{2}$

$\longrightarrow\rm {x}^{2} = \dfrac{ {a}^{2} }{2}$

$\longrightarrow \gray{\underline{ \boxed{\purple{\bold{ x = \pm \dfrac{ a}{ \sqrt{2} }}}} }}$

Hence,the option 1) x = ± (a/√2) is the correct answer.