JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 6)
Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius $$\mathrm{R}$$, with distance $$r$$ from the centre O is represented by:
_8th_April_Morning_Shift_en_6_1.png)
_8th_April_Morning_Shift_en_6_2.png)
_8th_April_Morning_Shift_en_6_3.png)
_8th_April_Morning_Shift_en_6_4.png)
_8th_April_Morning_Shift_en_6_5.png)
Explanation
Electric field due to the uniformly charged solid sphere is given by
$$ \begin{array}{ll} \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 r^2} \quad r \geq \mathrm{R} \\\\ \& & \\\\ \mathrm{E}=\frac{\mathrm{Q} r}{4 \pi \varepsilon_0 \mathrm{R}^3} \quad r \leq \mathrm{R} \end{array} $$
Therefore $\mathrm{E} \propto \mathrm{r}$ when
$\mathrm{r} \leq \mathrm{R}$ and $\mathrm{E} \propto \frac{1}{r^2}$ when $r \geq \mathrm{R}$
So
_8th_April_Morning_Shift_en_6_6.png)
$$ \begin{array}{ll} \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 r^2} \quad r \geq \mathrm{R} \\\\ \& & \\\\ \mathrm{E}=\frac{\mathrm{Q} r}{4 \pi \varepsilon_0 \mathrm{R}^3} \quad r \leq \mathrm{R} \end{array} $$
Therefore $\mathrm{E} \propto \mathrm{r}$ when
$\mathrm{r} \leq \mathrm{R}$ and $\mathrm{E} \propto \frac{1}{r^2}$ when $r \geq \mathrm{R}$
So
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