Electric Potential and Field Relationship

The problem asks to compute the Cartesian components of the electric field at a point, given the electric potential \(V\) at any point. The relationship between the electric field (\(\mathbf{E}\)) and the electric potential (\(V\)) is fundamental in electromagnetism. The electric field is the negative gradient of the electric potential.

Mathematically, this relationship is expressed as:

\[ \mathbf{E} = -\nabla V \]

In Cartesian coordinates, the gradient operator \(\nabla\) is given by:

\[ \nabla = \frac{\partial}{\partial x} \hat{\mathbf{i}} + \frac{\partial}{\partial y} \hat{\mathbf{j}} + \frac{\partial}{\partial z} \hat{\mathbf{k}} \]

Therefore, the Cartesian components of the electric field \((E_x, E_y, E_z)\) are calculated using partial derivatives of the potential \(V\) with respect to \(x\), \(y\), and \(z\):

• \( E_x = - \frac{\partial V}{\partial x} \)
• \( E_y = - \frac{\partial V}{\partial y} \)
• \( E_z = - \frac{\partial V}{\partial z} \)

Computing Electric Field Components

The given electric potential is \(V = x(y^2 - 4x^2)\). We can expand this expression to make differentiation easier:

\[ V = xy^2 - 4x^3 \]

Electric Field X-Component Calculation

To find the x-component of the electric field, \(E_x\), we need to take the partial derivative of \(V\) with respect to \(x\) and then negate it. When taking the partial derivative with respect to \(x\), we treat \(y\) as a constant.

First, let's find \( \frac{\partial V}{\partial x} \):

\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (xy^2 - 4x^3) \]

Differentiating term by term:

• The derivative of \(xy^2\) with respect to \(x\) is \(y^2\) (since \(y^2\) is treated as a constant).
• The derivative of \(-4x^3\) with respect to \(x\) is \(-4 \times 3x^{3-1} = -12x^2\).

So, we get:

\[ \frac{\partial V}{\partial x} = y^2 - 12x^2 \]

Now, to find \(E_x\), we negate this result:

\[ E_x = - (y^2 - 12x^2) = 12x^2 - y^2 \]

Electric Field Y-Component Calculation

To find the y-component of the electric field, \(E_y\), we take the partial derivative of \(V\) with respect to \(y\) and then negate it. When taking the partial derivative with respect to \(y\), we treat \(x\) as a constant.

First, let's find \( \frac{\partial V}{\partial y} \):

\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (xy^2 - 4x^3) \]

Differentiating term by term:

• The derivative of \(xy^2\) with respect to \(y\) is \(x \times 2y^{2-1} = 2xy\) (since \(x\) is treated as a constant).
• The derivative of \(-4x^3\) with respect to \(y\) is \(0\) (since \(-4x^3\) does not contain \(y\)).

So, we get:

\[ \frac{\partial V}{\partial y} = 2xy \]

Now, to find \(E_y\), we negate this result:

\[ E_y = - (2xy) = -2xy \]

Cartesian Components Summary

Finally, we need to consider the z-component of the electric field, \(E_z\). Since the potential \(V = xy^2 - 4x^3\) does not depend on \(z\), its partial derivative with respect to \(z\) will be zero.

\[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z} (xy^2 - 4x^3) = 0 \]

Therefore, \(E_z = -0 = 0\).

The Cartesian components of the electric field \((E_x, E_y)\) are:

\[ (E_x, E_y) = (12x^2 - y^2, -2xy) \]

This result provides the complete description of the electric field from the given potential function.