Question

Find the expression of electric potential within spherical shell between radius of r and r+dr​

Answer 1

Assume a solid conducting sphere of volume charge density $\rho.$ Consider a concentric sphere of radius $\sf{r}$ from it.

Then the charge enclosed by this shell will be,

$\sf{\longrightarrow q=\dfrac{4}{3}\,\pi\rho r^3}$

So the electric potential within this shell is,

$\sf{\longrightarrow V=\dfrac{kq}{r}}$

$\sf{\longrightarrow V=\dfrac{k\cdot\dfrac{4}{3}\,\pi\rho r^3}{r}}$

$\sf{\longrightarrow V=\dfrac{4}{3}\,k\pi\rho r^2}$

On differentiating this wrt $\sf{r,}$ we get the electric potential within a spherical shell between radius $\sf{r}$ and $\sf{r+dr.}$

$\sf{\longrightarrow dV=\dfrac{4}{3}\,k\pi\rho\cdot2r\,dr}$

$\sf{\longrightarrow\underline{\underline{dV=\dfrac{8}{3}\,k\pi\rho r\,dr}}}$

Taking $\sf{k=\dfrac{1}{4\pi\epsilon_0},}$

$\sf{\longrightarrow\underline{\underline{dV=\dfrac{2\rho r\,dr}{3\epsilon_0}}}}$

Answer 2

Answer:

Consider a concentric sphere of radius \sf{r}r from it.

Then the charge enclosed by this shell will be,

\sf{\longrightarrow q=\dfrac{4}{3}\,\pi\rho r^3}⟶q=34πρr3

So the electric potential within this shell is,

\sf{\longrightarrow V=\dfrac{kq}{r}}⟶V=rkq

\sf{\longrightarrow V=\dfrac{k\cdot\dfrac{4}{3}\,\pi\rho r^3}{r}}⟶V=rk⋅34πρr3

\sf{\longrightarrow V=\dfrac{4}{3}\,k\pi\rho r^2}⟶V=34kπρr2

On differentiating this wrt \sf{r,}r, we get the electric potential within a spherical shell between radius \sf{r}r and \sf{r+dr.}r+dr.

\sf{\longrightarrow dV=\dfrac{4}{3}\,k\pi\rho\cdot2r\,dr}⟶dV=34kπρ⋅2rdr

\sf{\longrightarrow\underline{\underline{dV=\dfrac{8}{3}\,k\pi\rho r\,dr}}}⟶dV=38kπρrdr

Taking \sf{k=\dfrac{1}{4\pi\epsilon_0},}k=4πϵ01,

\sf{\longrightarrow\underline{\underline{dV=\dfrac{2\rho r\,dr}{3\epsilon_0}}}}⟶dV=3ϵ02ρrdr

Step-by-step explanation:

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