MHT CET · Physics · Electrostatics
The electric field intensity on the surface of a solid charged sphere of radius ' \(r\) ' and volume charge density ' \(\rho\) ' is ( \(\varepsilon_0=\) permittivity of free space \()\)
- A \(\frac{\rho r}{3 \varepsilon_0}\)
- B \(\frac{\rho}{4 \pi \varepsilon_0 r}\)
- C zero
- D \(\frac{5 \rho \mathrm{r}}{6 \varepsilon_0}\)
Answer & Solution
Correct Answer
(A) \(\frac{\rho r}{3 \varepsilon_0}\)
Step-by-step Solution
Detailed explanation
According to Gauss' theorem,
\(
\phi=\mathrm{E} \cdot \mathrm{A}=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_0}
\)
But, \(\mathrm{q}_{\text {enc }}=\rho \times\) volume
\(
\therefore \mathrm{q}_{\mathrm{enc}}=\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)
\)
And area is \(A=4 \pi r^2\)
\(\therefore \mathrm{E}\left(4 \pi \mathrm{r}^2\right)=\frac{\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)}{\varepsilon_0}\)
\(\therefore \mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}<\)
\(
\phi=\mathrm{E} \cdot \mathrm{A}=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_0}
\)
But, \(\mathrm{q}_{\text {enc }}=\rho \times\) volume
\(
\therefore \mathrm{q}_{\mathrm{enc}}=\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)
\)
And area is \(A=4 \pi r^2\)
\(\therefore \mathrm{E}\left(4 \pi \mathrm{r}^2\right)=\frac{\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)}{\varepsilon_0}\)
\(\therefore \mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}<\)
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