Question

The potential (in volts) of a charge distribution is given by
   $$V\left(z\right)=30-5z^{2}for\left|z\right|⩽1m$$
    $$V\left(z\right)=35-10zfor\left|z\right|⩾1m$$
    $$V\left(z\right)doesnotdependsonxandy.$$
If the potential is generated by a constant charge per unit volume $$r_0(inunitsof{e}_0)$$which is
 spread over a certain region, then choose the correct statement:
 A. $$r_0=20e_{0}for\left|z\right|⩽1mand{r}_0=0$$ elsewhere.
 B. $$.r_0=40e_0$$ in the entire region.
 C. $$r_0=10e_{0}for\left|z\right|⩽1mand{r}_0=0$$ elsewhere.
 D. $$r_0=20e_0$$ in the entire region.

Answer 1

Hint:
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
${}^{{\mathrm{\nabla }}^{2}V=-{\displaystyle \frac{\rho }{{\epsilon }_0}}}$
3. Every electrical potential follows Poisson’s equation.

Complete step by step solution :
For $$\left|z\right|⩽1m$$,
 $$V\left(z\right)=30-5z^2\left(given\right).$$
Now, Putting This expression of V(z) in Poisson’s equation we get,
$$\begin{gathered} {}^{{\mathrm{\nabla }}^{2}V=-{\displaystyle \frac{\rho }{{\epsilon }_0}}} \\ \Rightarrow {\displaystyle \frac{{\partial }^{2}V}{\partial x^2}}+{\displaystyle \frac{{\partial }^{2}V}{\partial y^2}}+{\displaystyle \frac{{\partial }^{2}V}{\partial z^2}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow {\displaystyle \frac{{\partial }^{2}V}{\partial z^2}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow {\displaystyle \frac{{\partial }^2(30-5z^2)}{\partial z^2}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow {\displaystyle \frac{\partial (-10z)}{\partial z}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow -10=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow \rho =10{\epsilon }_0 \end{gathered}$$
As V is independent of x and y, so the $$1^{st}$$two terms of $${\mathrm{\nabla }}^{2}V$$ vanishes.
So for $$\left|z\right|⩽1m$$ region the charge density is$10{\epsilon }_0$
Now, For $$\left|z\right|⩾1m,V\left(z\right)=35-10z\left(given\right).$$
Putting This expression of $$V\left(z\right)$$in Poisson’s equation we get,
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\partial }^2(35-10z)}{\partial z^2}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow {\displaystyle \frac{\partial (-10)}{\partial z}}=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow 0=-{\displaystyle \frac{\rho }{{\epsilon }_0}} \\ \Rightarrow \rho =0 \end{gathered}$$
  So for $$\left|z\right|⩾1m$$ region the charge density is zero.
Therefore the final answer is $$r_0=10e_{0}for\left|z\right|⩽1mand{r}_0=0$$ 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 $$\left(inwhichcoordinatesystemitisgiven\right)$$because for cartesian coordinate system the expression of $${\mathrm{\nabla }}^{2}V$$ is given in solution but for cylindrical and spherical polar Coordinate system the expression of $${\mathrm{\nabla }}^{2}V$$ is different.
2. Extra care should be given while doing the partial derivative.