Let rho (r), = frac{Q}{{pi {R^4}}},r be | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{dq}=\frac{\mathrm{Qr}}{\pi \mathrm{R}^{4}} 	imes 4 \pi \mathrm{r}^{2} \cdot \mathrm{dr}$

${m{q}} = \frac{{4{m{Q}}}}{{{{m{R}}^4}}}\i

Vedclass · Accepted Answer

$\frac{{Qr_{_1}^2}}{{4\pi {\varepsilon _0}{R^4}}}\,$

Vedclass · Answer

$\mathrm{dq}=\frac{\mathrm{Qr}}{\pi \mathrm{R}^{4}} 	imes 4 \pi \mathrm{r}^{2} \cdot \mathrm{dr}$

${m{q}} = \frac{{4{m{Q}}}}{{{{m{R}}^4}}}\int_0^{{{m{r}}_1}} {{{m{r}}^3}} \quad {m{dr}} = \frac{{{m{Qr}}_1^4}}{{{{m{R}}^4}}}$

Now the electric field at a distance $\mathrm{r}_{1}$ from the centre (inside)

$\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}_{1}^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Qr}_{1}^{2}}{\mathrm{R}^{4}}$

Let $\rho (r)\, = \frac{Q}{{\pi {R^4}}}\,r$ be the volume charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $'p'$ inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

Similar Questions

According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to

Explain by graph how the electric field by thin spherical shell depends on the distance of point from centre.

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{ I }, E_{ II }$ and $E_{III}$ are

Consider $a$ uniformly charged hemispherical shell of radius $R$ and charge $Q$ . If field at point $A (0, 0, -z_0)$ is $ \vec E$ then field at point $(0, 0, z_0)$ is $[z_0 < R]$