एक गोलीय सममिती आवेश वितरण आवेश घनत्व का निम्नलिखित विचरण रखता है : \(\rho(r)=\rho_{o}\left(1-\frac{r}{R}\right) r < R\) के लिए \(\rho( r )=0 \quad r \geqslant R\) के लिए जहाँ \(r\) आवेश वितरण के केन्द्र से दूरी हैं और \(\rho_{ o }\) एक स्थिरांक है। एक अन्तः बिन्दु \(( r < R )\) पर विद्युत क्षेत्र है

A \(\frac{{{\rho _0}}}{{4{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)
B \(\frac{{{\rho _0}}}{{{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)
C \(\frac{{{\rho _0}}}{{3{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)
D \(\frac{{{\rho _0}}}{{12{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{{{\rho _0}}}{{{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)

Step-by-step Solution

Detailed explanation

Let us consider a spherical shell of radius \(x\) and thickness \(dx.\) Charge on this shell \(\mathrm{dq}=\rho 4 \pi \mathrm{x}^{2} \mathrm{dx}=\rho_{0}\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right) .4 \pi \mathrm{x}^{2} \mathrm{dx}\) \(\therefore\) Total charge in the spherical…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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