The potential (in volts) of a charge distribution is given by
$$V(z) = 30 - 5z^2$$ for $$|z| \leq 1$$ m
$$V(z) = 35 - 10|z|$$ for $$|z| \geq 1$$ m.
$$V(z)$$ does not depend on x and y. If this potential is generated by a constant charge per unit volume $$\rho_0$$ (in units of $$\epsilon_0$$) which is spread over a certain region, then choose the correct statement.

To link the given electric potential to the corresponding volume charge density we recall the fundamental relation from electrostatics, namely Poisson’s equation, stated in SI units as

$$\nabla^{2}V = -\,\frac{\rho}{\epsilon_{0}}.$$

Because the problem tells us that the potential $$V$$ depends only on the coordinate $$z$$ (it is uniform in the $$x$$ and $$y$$ directions), the Laplacian simplifies dramatically. In Cartesian coordinates, with no $$x$$ or $$y$$ dependence, we have

$$\nabla^{2}V \;=\; \frac{\partial^{2}V}{\partial x^{2}} + \frac{\partial^{2}V}{\partial y^{2}} + \frac{\partial^{2}V}{\partial z^{2}} \;=\; 0 + 0 + \frac{d^{2}V}{dz^{2}} \;=\; \frac{d^{2}V}{dz^{2}}.$$

So Poisson’s equation for this one-dimensional situation becomes

$$\frac{d^{2}V}{dz^{2}} = -\,\frac{\rho(z)}{\epsilon_{0}}.$$

Now we examine the potential piece-wise.

Region 1 : $$|z|\le 1\text{ m}$$

The potential is given by $$V(z)=30-5z^{2}.$$ Let us differentiate twice:

First derivative: $$\frac{dV}{dz} = -10\,z.$$

Second derivative: $$\frac{d^{2}V}{dz^{2}} = -10.$$

This second derivative is a constant, so in this region

$$\nabla^{2}V = -10.$$

Substituting into Poisson’s equation,

$$-10 = -\,\frac{\rho}{\epsilon_{0}} \;\;\Longrightarrow\;\; \rho = 10\,\epsilon_{0}.$$

Thus the volume charge density inside the slab $$|z|\le 1\text{ m}$$ is a positive and uniform value $$\rho_0 = 10\,\epsilon_{0}.$$

Region 2 : $$|z|\ge 1\text{ m}$$

The potential now is given by $$V(z)=35-10|z|.$$ We treat the two sides separately but the mathematical outcome is the same.

For $$z\ge 1$$ we have $$|z|=z,$$ so $$V(z)=35-10z.$$ For $$z\le -1$$ we have $$|z|=-z,$$ so $$V(z)=35+10z.$$

Differentiating either expression yields a constant first derivative, and the second derivative is zero:

$$\frac{d^{2}V}{dz^{2}} = 0 \quad(\text{for }|z|\ge 1).$$

Therefore, in this exterior region

$$\nabla^{2}V = 0 \;\;\Longrightarrow\;\; 0 = -\,\frac{\rho}{\epsilon_{0}} \;\;\Longrightarrow\;\; \rho = 0.$$

The only non-zero volume charge resides in the central slab $$|z|\le 1\text{ m},$$ and its magnitude is $$10\,\epsilon_{0}.$$ Outside that slab the charge density is identically zero. (The kink in the slope of $$V$$ at $$|z|=1$$ would correspond to an additional surface charge, but the question asks specifically for a volume charge of constant magnitude, so that surface effect is not relevant to the choices given.)

Comparing this result with the listed options, we see that it matches exactly with Option B.

Hence, the correct answer is Option B.