Electric potential in a region is varying according to the relation V = `(3x^(2))/(2) - (y^(2))/(4)`, where x and y are in metre and V is volt, Electric field intensity (in N/C) at a point (1 m, 2 m ) is

Options

A. `3 hat(i) - hat(j)`
B. `- 3 hat(i) + hat(j)`
C. `6 hat(i) - 2hat(j)`
D. `- 6 hat(i) + 2hat(j)`

Answer

To find the electric field intensity at the point (1 m, 2 m) given the electric potential \( V = \frac{3x^2}{2} - \frac{y^2}{4} \), we will follow these steps:

### Step 1: Understand the relationship between electric potential and electric field
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
This means that the electric field can be found by taking the negative gradient of the electric potential.

### Step 2: Calculate the x-component of the electric field
The x-component of the electric field \( E_x \) is given by:
\[ E_x = -\frac{\partial V}{\partial x} \]
First, we need to differentiate \( V \) with respect to \( x \):
\[ V = \frac{3x^2}{2} - \frac{y^2}{4} \]
Differentiating \( V \) with respect to \( x \) (treating \( y \) as a constant):
\[ \frac{\partial V}{\partial x} = \frac{3}{2} \cdot 2x = 3x \]
Thus, the x-component of the electric field is:
\[ E_x = -3x \]
Now, substituting \( x = 1 \) m:
\[ E_x = -3(1) = -3 \, \text{N/C} \]

### Step 3: Calculate the y-component of the electric field
The y-component of the electric field \( E_y \) is given by:
\[ E_y = -\frac{\partial V}{\partial y} \]
Now, we differentiate \( V \) with respect to \( y \) (treating \( x \) as a constant):
\[ \frac{\partial V}{\partial y} = -\frac{y}{2} \]
Thus, the y-component of the electric field is:
\[ E_y = -\left(-\frac{y}{2}\right) = \frac{y}{2} \]
Now, substituting \( y = 2 \) m:
\[ E_y = \frac{2}{2} = 1 \, \text{N/C} \]

### Step 4: Combine the components to find the electric field vector
Now we can combine the x and y components to find the total electric field vector at the point (1 m, 2 m):
\[ \vec{E} = E_x \hat{i} + E_y \hat{j} = -3 \hat{i} + 1 \hat{j} \]

### Final Answer
The electric field intensity at the point (1 m, 2 m) is:
\[ \vec{E} = -3 \hat{i} + 1 \hat{j} \, \text{N/C} \]

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