Consider a solid insulating sphere of radius | vedclass

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Question

${{m{E}}_{m{x}}} 	imes 4\pi {{m{x}}^2} = \frac{{{ho _0}\int_0^{m{x}} {{{m{x}}^2}}  	imes 4\pi {{m{x}}^2}{m{dx}}}}{{{ \in _0}}}$

Vedclass · Accepted Answer

${x^3}y^2 = {R^5}$

Vedclass · Answer

${{m{E}}_{m{x}}} 	imes 4\pi {{m{x}}^2} = \frac{{{ho _0}\int_0^{m{x}} {{{m{x}}^2}}  	imes 4\pi {{m{x}}^2}{m{dx}}}}{{{ \in _0}}}$

$ = \frac{{4\pi {ho _0}{x^5}}}{{5{ \in _0}}}$

${{m{E}}_y} = \frac{{{ho _0}\int_0^{m{R}} {{{m{x}}^2}}  	imes 4\pi {{m{x}}^2}{m{dx}}}}{{{ \in _{m{u}}} 	imes 4\pi {{m{y}}^2}}} = \frac{{{ho _0} 	imes {{m{R}}^5}}}{{5{ \in _0}{{m{y}}^2}}}$

$\frac{{{ho _0}{{m{x}}^3}}}{{5{ \in _0}}} = \frac{{{ho _0}{{m{R}}^5}}}{{5{ \in _0}{{m{y}}^2}}}$

$\mathrm{x}^{3} \mathrm{y}^{2}=\mathrm{R}^{5}$

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