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If the electrostatic potential in spherical polar coordinates is  \phi(r)=\phi e^{-r / r_0}  where \phi  and r_0  are constant, then the charge density at a distance r=r_0  will be-

Option: 1

\frac{\epsilon_0 \phi_0}{e_0^2}


Option: 2

\frac{e \epsilon_0 \phi_0}{2 \gamma_0^2}


Option: 3

-\frac{\epsilon_0 \phi_0}{e \gamma_0^2}


Option: 4

-\frac{2 e \epsilon_0 \psi_0}{\gamma_0^2}


Answers (1)

best_answer

Poission's equation,    \nabla^2 \phi=-\frac{\rho}{\epsilon_0} \Rightarrow \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)


\Rightarrow \frac{1}{r^2} \frac{\partial}{\partial r}\left[r^2 \frac{\partial}{\partial r}\left(\phi e^{-r / r_0}\right)\right]=-\frac{\rho}{\epsilon_0} \\


-\frac{\phi_0}{r_0 r^2} \frac{\partial}{\partial r}\left(r^2 e^{-r / r_0}\right)=-\frac{\rho}{\\\varepsilon _0} \\


\frac{\phi_0}{r_0 r^2}\left[2 r e^{-r / r_0}-\frac{r^2}{r_0} e^{-r/ r_0}\right]=\frac{\rho}{\epsilon }

At r=r_0, 

\frac{\phi_0}{r_0^3}[2r_0e^{-1}-r_0 e^{-1}]=\frac{\rho}{\epsilon_0}

\frac{\phi_0e^{-1}}{r_0^2}=\frac{\rho}{\epsilon_0}

\rho=\frac{\phi_0\epsilon_0}{r_0^2e}

 

 

Posted by

shivangi.shekhar

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