The potential (in volts) of a charge distribution is given by V(z)=30−5z2V(z)=30−5z2 for |z|≤1m|z|≤1mV(z)=35−10|z|V(z)=35−10|z| for |z|≥1m|z|≥1m V(z)V(z) does not depend on x and y .If this potential is generated by a constant charge per unit volume ρ0,ρ0, (in units of ∈0∈0) which is spread over certain region ,then choose the correct statement

To analyze the potential distribution you've provided, we need to understand how electric potential relates to charge density. The potential \( V(z) \) is defined in two regions, and we can derive the charge density \( \rho_0 \) from it using Gauss's law and the relationship between electric field and potential.

Understanding the Potential Function

The potential function is given as:

• For \( |z| \leq 1 \, \text{m} \): \( V(z) = 30 - 5z^2 \)
• For \( |z| \geq 1 \, \text{m} \): \( V(z) = 35 - 10|z| \)

Finding the Electric Field

The electric field \( E \) is related to the potential \( V \) by the equation:

E(z) = -\frac{dV}{dz}

Let's calculate the electric field in both regions:

For \( |z| \leq 1 \, \text{m} \):

Taking the derivative:

E(z) = -\frac{d}{dz}(30 - 5z^2) = 10z

For \( |z| \geq 1 \, \text{m} \):

Again, taking the derivative:

E(z) = -\frac{d}{dz}(35 - 10|z|) = -10 \, \text{sgn}(z)

Here, \( \text{sgn}(z) \) is the sign function, which is +1 for \( z > 0 \) and -1 for \( z < 0 \).

Relating Electric Field to Charge Density

According to Gauss's law, the divergence of the electric field relates to the charge density:

\(\nabla \cdot E = \frac{\rho}{\epsilon_0}\)

In one dimension, this simplifies to:

\(\frac{dE}{dz} = \frac{\rho}{\epsilon_0}\)

Calculating Charge Density

Now, let's find the charge density in both regions:

For \( |z| \leq 1 \, \text{m} \):

We have:

\(\frac{dE}{dz} = \frac{d(10z)}{dz} = 10\)

Thus, the charge density is:

\(\rho = 10 \epsilon_0\)

For \( |z| \geq 1 \, \text{m} \):

Here, the electric field is constant:

\(\frac{dE}{dz} = 0\)

This implies:

\(\rho = 0\)

Summarizing the Findings

From our calculations, we can conclude:

• In the region \( |z| \leq 1 \, \text{m} \), there is a uniform charge density of \( 10 \epsilon_0 \).
• In the region \( |z| \geq 1 \, \text{m} \), there is no charge density, indicating that the potential is influenced by the charge distribution only within the first meter.

Therefore, the correct statement regarding the charge distribution is that it is a constant charge per unit volume \( \rho_0 = 10 \epsilon_0 \) in the region \( |z| \leq 1 \, \text{m} \) and zero elsewhere. This understanding of the relationship between electric potential, electric field, and charge density is fundamental in electrostatics and helps us visualize how charges influence the space around them.