Let ( r )= Q R ^ 4 r be the charge density distribution for a solid sphere of radius R…
Physics · JEE Main · NTA Exams — Electrostatics
Let \(\rho(\mathrm{r})=\frac{\mathrm{Q}}{\pi \mathrm{R}^{4}} \mathrm{r}\)be the charge density distribution for a solid sphere of radius R and total charge Q. For a point P inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is
- zero
- \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{r}_{1}^{2}}\)
- \(\frac{\mathrm{Qr}_{1}^{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{4}}\)
- \(\frac{\mathrm{Qr}_{1}^{2}}{3 \pi \varepsilon_{0} \mathrm{R}^{4}}\)
Answer
(C) Qr _ 1 ^ 2 4 _ 0 R ^ 4
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