The potential in certain region is given as `V = 2x^(2)`, then the charge density of that region is

Options

A. `-(4x)/(epsilon_(0))`
B. `-(4)/(epsilon_(0))`
C. `-4 epsilon_(0)`
D. `-2 epsilon_(0)`

Answer

To find the charge density in a region where the electric potential \( V \) is given by \( V = 2x^2 \), we can use the relationship between electric potential and charge density derived from Gauss's law. The relevant equation is:

\[ \nabla^2 V = -\frac{\rho}{\epsilon_0} \]

where \( \nabla^2 \) is the Laplacian operator, \( \rho \) is the charge density, and \( \epsilon_0 \) is the permittivity of free space.

### Step 1: Calculate the Laplacian of the potential \( V \)

Given the potential \( V = 2x^2 \), we first need to compute the second derivative of \( V \) with respect to \( x \):

\[ \frac{dV}{dx} = \frac{d}{dx}(2x^2) = 4x \]

Now, we take the second derivative:

\[ \frac{d^2V}{dx^2} = \frac{d}{dx}(4x) = 4 \]

### Step 2: Apply the Laplacian to find \( \nabla^2 V \)

In one dimension, the Laplacian \( \nabla^2 V \) simplifies to:

\[ \nabla^2 V = \frac{d^2V}{dx^2} = 4 \]

### Step 3: Relate the Laplacian to charge density

Now, we substitute this result into the equation relating the Laplacian to charge density:

\[ \nabla^2 V = -\frac{\rho}{\epsilon_0} \]

Substituting \( \nabla^2 V = 4 \):

\[ 4 = -\frac{\rho}{\epsilon_0} \]

### Step 4: Solve for charge density \( \rho \)

Rearranging the equation to solve for \( \rho \):

\[ \rho = -4 \epsilon_0 \]

### Final Answer

Thus, the charge density in the region is:

\[ \rho = -4 \epsilon_0 \]

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