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Question

${E_r}\,\, = \,\, - \,\,\frac{{\partial \varphi }}{{\partial r}}\,\, = \,\, - 2ar$

ગોસ પ્રમેય પરથી $4\pi {r^2}{E_r}\, = \,\,\frac{q}{{{\varepsilon

Vedclass · Accepted Answer

$-6\, a$$\varepsilon_0$

Vedclass · Answer

${E_r}\,\, = \,\, - \,\,\frac{{\partial \varphi }}{{\partial r}}\,\, = \,\, - 2ar$

ગોસ પ્રમેય પરથી $4\pi {r^2}{E_r}\, = \,\,\frac{q}{{{\varepsilon _0}}}  $

વિકલન કરતાં $4\pi d\,\,\left( {{r^2}{E_r}} \right)\,\, = \,\,\frac{{dq}}{{{\varepsilon _0}}}\,\, = \,\,\frac{{\left( {4\pi {e^2}dr} \right)\,\rho }}{{{\varepsilon _0}}}$

$ \Rightarrow \,\,{r^2}d{E_r}\,\, + \;\,2r{E_r}\,\,dr\,\, = \,\,\frac{1}{{{\varepsilon _0}}}\,\rho {r^2}dr\,\,\, \Rightarrow \,\frac{{\partial {E_r}}}{{\partial r}}\,\, + \,\,\frac{2}{r}\,\,{E_r}\,\, = \,\,\frac{\rho }{{{\varepsilon _0}}}\,\, \Rightarrow \,\,\rho \,\, = \,\, - 6{\varepsilon _0}a$

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