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JEE Mains · Physics · STD 12 - 1. Electric charges and fields

\(R\) ત્રિજ્યા ધરાવતા ઘન ગોળની વિજભાર ઘનતા \(0 \leq r \leq R\) માટે \(\rho  = {\rho _0}\left( {1 - \frac{r}{R}} \right)\) મુજબ આપવામાં આવે છે. તો બોલની બહાર વિદ્યુતક્ષેત્ર કેટલું હશે?

  1. A \(\frac{{{\rho _0}{R^3}}}{{{\varepsilon _0}{r^2}}}\)
  2. B \(\frac{{{4\rho _0}{R^3}}}{{{3\varepsilon _0}{r^2}}}\)
  3. C \(\frac{{{3\rho _0}{R^3}}}{{{4\varepsilon _0}{r^2}}}\)
  4. D \(\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}\)

Step-by-step Solution

Detailed explanation

\(\text { Charge density, } \rho=\rho_{0}\left(1-\frac{r}{R}\right)\) \(d q=\rho d v\) \(q_{i n}=\int d q=\rho d v\) \(=\rho_{0}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r \quad\left(\because d v=4 \pi \mathrm{r}^{2} \mathrm{dr}\right)\)…
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