Question

The potential (in volts) of a charge distribution is given by
$V\left( z \right){\text{ }} = {\text{ }}30{\text{ }} - {\text{ }}5{z^2}for{\text{ }}\left| z \right| \leqslant 1m$
$V\left( z \right){\text{ }} = {\text{ }}35{\text{ }} - {\text{ }}10z{\text{ }}for{\text{ }}\left| z \right| \geqslant 1m$
$V\left( z \right){\text{ }}does{\text{ }}not{\text{ }}depends{\text{ }}on{\text{ }}x{\text{ }}and{\text{ }}y.$
If the potential is generated by a constant charge per unit volume ${r_0}(in{\text{ }}units{\text{ }}of{e_0})$which is
spread over a certain region, then choose the correct statement:
A. ${r_0} = {\text{ }}20{e_0}for{\text{ }}\left| z \right| \leqslant 1m{\text{ }} and {r_0} = 0$ elsewhere.
B. $.{r_0} = 40{e_0}$ in the entire region.
C. ${r_0} = {\text{ }}10{e_0}for{\text{ }}\left| z \right| \leqslant 1m{\text{ }} and {r_0} = 0$ elsewhere.
D. ${r_0} = 20{e_0}$ in the entire region.

Answer 1

1. Here before solving this problem we need to know about Gauss’ law and also Poisson’s equation.
2. If V is the electrical potential generated by a static charge distribution of charge density ρ then according to Poisson’s equation
$^{{\nabla ^2}V = - \dfrac{\rho }{{{\varepsilon _0}}}}$
3. Every electrical potential follows Poisson’s equation.

Complete step by step solution :
For $\left| z \right| \leqslant 1m$,
$V\left( z \right) = {\text{ }}30{\text{ }} - {\text{ }}5{z^2}\left( {given} \right).$
Now, Putting This expression of V(z) in Poisson’s equation we get,

$$^{{\nabla ^2}V = - \dfrac{\rho }{{{\varepsilon _0}}}} \\\ \Rightarrow \dfrac{{{\partial ^2}V}}{{\partial {x^2}}} + \dfrac{{{\partial ^2}V}}{{\partial {y^2}}} + \dfrac{{{\partial ^2}V}}{{\partial {z^2}}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \dfrac{{{\partial ^2}V}}{{\partial {z^2}}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \dfrac{{{\partial ^2}(30 - 5{z^2})}}{{\partial {z^2}}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \dfrac{{\partial ( - 10z)}}{{\partial z}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow - 10 = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \rho = 10{\varepsilon _0} \\\$$

As V is independent of x and y, so the ${1^{st}}$two terms of ${\nabla ^2}V$ vanishes.
So for $\left| z \right| \leqslant 1m$ region the charge density is$10{\varepsilon _0}$
Now, For $\left| z \right| \geqslant 1m,{\text{ }}V\left( z \right) = {\text{ }}35 - 10z{\text{ }}\left( {given} \right).$
Putting This expression of $V\left( z \right)$in Poisson’s equation we get,
$\\\ \Rightarrow \dfrac{{{\partial ^2}(35 - 10z)}}{{\partial {z^2}}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \dfrac{{\partial ( - 10)}}{{\partial z}} = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow 0 = - \dfrac{\rho }{{{\varepsilon _0}}} \\\ \Rightarrow \rho = 0 \\\$
So for $\left| z \right| \geqslant 1m$ region the charge density is zero.
Therefore the final answer is ${r_0} = {\text{ }}10{e_0}for{\text{ }}\left| z \right| \leqslant 1m{\text{ }} and {r_0} = 0$ elsewhere.
Hence the correct answer is (C).

Note:
1. In this type of problem we have to use proper expression of gradient operator based on the potential $\left( {in{\text{ }}which{\text{ }}coordinate{\text{ }}system{\text{ }}it{\text{ }}is{\text{ }}given} \right)$because for cartesian coordinate system the expression of ${\nabla ^2}V$ is given in solution but for cylindrical and spherical polar Coordinate system the expression of ${\nabla ^2}V$ is different.
2. Extra care should be given while doing the partial derivative.