The electirc potential at a point `(x, y, z)` is given by
`V = -x^(2)y - xz^(3) + 4`
The electric field `vecE` at that point is

Options

A. `E =hat(i)(2xy+z^(3))+hat(j)x^(2)+3xz^(2)hat(k)`
B. `E=hat(i)2xy+hat(i)(x^(2)+y^(2))+hat(k)(3xz-y^(2))`
C. `E=hat(i)z^(3)+hat(j)xyz+hat(k)z^(2)`
D. `E=hat(i)(2xy-z^(3))+hat(j)xy^(2)+hat(k)3z^(2)x`

Answer

To find the electric field \(\vec{E}\) at the point \((x, y, z)\) given the electric potential \(V = -x^2y - xz^3 + 4\), we will use the relationship between electric potential and electric field. The electric field is defined as:

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

where \(\nabla V\) is the gradient of the potential \(V\). The gradient in three dimensions is given by:

\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]

### Step 1: Calculate \(\frac{\partial V}{\partial x}\)

We differentiate \(V\) with respect to \(x\):

\[ V = -x^2y - xz^3 + 4 \]

Taking the partial derivative with respect to \(x\):

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

### Step 2: Calculate \(\frac{\partial V}{\partial y}\)

Next, we differentiate \(V\) with respect to \(y\):

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

### Step 3: Calculate \(\frac{\partial V}{\partial z}\)

Now, we differentiate \(V\) with respect to \(z\):

\[ \frac{\partial V}{\partial z} = -3xz^2 \]

### Step 4: Write the gradient of \(V\)

Now we can write the gradient of \(V\):

\[ \nabla V = \left(-2xy - z^3, -x^2, -3xz^2\right) \]

### Step 5: Calculate the electric field \(\vec{E}\)

Using the relationship \(\vec{E} = -\nabla V\):

\[ \vec{E} = -\left(-2xy - z^3, -x^2, -3xz^2\right) \]

This simplifies to:

\[ \vec{E} = \left(2xy + z^3, x^2, 3xz^2\right) \]

### Final Answer

Thus, the electric field \(\vec{E}\) at the point \((x, y, z)\) is:

\[ \vec{E} = (2xy + z^3) \hat{i} + (x^2) \hat{j} + (3xz^2) \hat{k} \]

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