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 follow these steps:

### Step 1: Write down the expression for electric potential
The electric potential \( V \) is given by:
\[ V = 9(x^3 - y^3) \]

### Step 2: Recall the relationship between electric field and potential
The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the formula:
\[ \mathbf{E} = -\nabla V \]
In Cartesian coordinates, this can be expressed as:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) \]

### Step 3: Calculate the partial derivative of \( V \) with respect to \( x \)
To find the \( x \)-component of the electric field \( E_x \):
\[ E_x = -\frac{\partial V}{\partial x} \]
Calculating the derivative:
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(9(x^3 - y^3)) = 9 \cdot 3x^2 = 27x^2 \]
Thus,
\[ E_x = -27x^2 \]

### Step 4: Calculate the partial derivative of \( V \) with respect to \( y \)
To find the \( y \)-component of the electric field \( E_y \):
\[ E_y = -\frac{\partial V}{\partial y} \]
Calculating the derivative:
\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(9(x^3 - y^3)) = 9 \cdot (-3y^2) = -27y^2 \]
Thus,
\[ E_y = -(-27y^2) = 27y^2 \]

### Step 5: Combine the components to write the electric field vector
Now we can write the electric field vector \( \mathbf{E} \):
\[ \mathbf{E} = (E_x, E_y) = (-27x^2, 27y^2) \]

### Step 6: Final expression for the electric field strength vector
Thus, the electric field strength vector is:
\[ \mathbf{E} = -27x^2 \hat{i} + 27y^2 \hat{j} \]