Potential of electric ield vector is given by V= 9 `(x^3 - y^3 )`, (where , xand y are cartesian Co-ordinates). The electric field strenth vector is

Options

A. `-27x^2 hat i + 27y^2 hat j`
B. `27x hat i + 27y hat j`
C. `27x^2 y hat i + 27xy^2 hat j`
D. `-27xy hat i + 27xy hat j`

Answer

To find the electric field strength vector \( \mathbf{E} \) from the given electric potential \( V \), we can use the relationship between electric potential and electric field. The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the equation:

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

where \( \nabla V \) is the gradient of the potential \( V \). In Cartesian coordinates, this can be expressed as:

\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right) \]

Given the potential:

\[ V = 9(x^3 - y^3) \]

we will calculate the partial derivatives with respect to \( x \) and \( y \).

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

\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(9(x^3 - y^3)) = 9 \cdot \frac{\partial}{\partial x}(x^3) - 9 \cdot \frac{\partial}{\partial x}(y^3) \]

Since \( y \) is treated as a constant when differentiating with respect to \( x \):

\[ \frac{\partial V}{\partial x} = 9 \cdot 3x^2 = 27x^2 \]

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

\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(9(x^3 - y^3)) = 9 \cdot \frac{\partial}{\partial y}(x^3) - 9 \cdot \frac{\partial}{\partial y}(y^3) \]

Since \( x \) is treated as a constant when differentiating with respect to \( y \):

\[ \frac{\partial V}{\partial y} = -9 \cdot 3y^2 = -27y^2 \]

### Step 3: Substitute into the electric field equation

Now we can substitute these derivatives into the equation for the electric field:

\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right) \]

Substituting the values we found:

\[ \mathbf{E} = -\left( 27x^2 \hat{i} - 27y^2 \hat{j} \right) \]

This simplifies to:

\[ \mathbf{E} = -27x^2 \hat{i} + 27y^2 \hat{j} \]

### Final Result

Thus, the electric field strength vector is:

\[ \mathbf{E} = -27x^2 \hat{i} + 27y^2 \hat{j} \]

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