Question

If the potential given by $\overrightarrow{V}=\frac{20sin\theta }{r^3}$ in a region, then value of volume charge density ${\rho }_v$ at point $P(r=2,\theta ={30}^{\circ },\varphi =0^{\circ })$ will be

• A. $84.21pC/m^3$
• B. $-20.24pC/m^3$
• C. $80.42pC/m^3$
• D. $-22.12pC/m^3$

Answer 1

The correct option is D $-22.12pC/m^3$

By Poisson equation
${\nabla }^{2}V=\frac{-{\rho }_v}{{ϵ}_0}$
As V is not a function of $\varphi ,{\nabla }^{2}V$ in the spherical coordinates reduces to
${\nabla }^{2}V=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial V}{\partial \theta }\right)$

$=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\left(\frac{20sin\theta }{r^3}\right)\right)+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }(sin\theta .\frac{20cos\theta }{r^3})$

$=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2.\frac{-60sin\theta }{r^4})+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }(sin\theta .\frac{20cos\theta }{r^3})$

$=\frac{1}{r^2}\frac{\partial }{\partial r}\left(\frac{-60sin\theta }{r^2}\right)+\frac{1}{r^{2}sin\theta }\frac{\partial }{\partial \theta }\left(\frac{10sin2\theta }{r^3}\right)$

$=\frac{1}{r^2}\times \left(\frac{120sin\theta }{r^3}\right)+\frac{1}{r^{2}sin\theta }\frac{20cos2\theta }{r^3}$

${\nabla }^{2}V=\frac{120sin\theta }{r^5}+\frac{20cos2\theta }{r^{5}sin\theta }$

$=\frac{20}{r^{5}sin\theta }[4si{n}^2\theta +1]$

Now, ${\nabla }^{2}V=\frac{-{\rho }_v}{{ϵ}_0}$

${\rho }_v=-{ϵ}_0\left[\frac{20(4si{n}^2\theta +1)}{r^{5}sin\theta }\right]$

At $P(2,{30}^{\circ },0^{\circ })$ ${\rho }_v=-{ϵ}_0\left[\frac{20(4si{n}^2{30}^{\circ }+1)}{2^{5}sin{30}^{\circ }}\right]$

$=-{ϵ}_0\left[\frac{40}{32\times 0.5}\right]$

$=-22.12\times {10}^{-12}C/m^3$

$=-22.12pC/m^3$