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 \( \mathbf{E} \) is related to the electric potential \( V \) by the equation:
\[ \mathbf{E} = -\nabla V \]
where \( \nabla V \) is the gradient of the potential.

### Step 2: Calculate the partial derivatives of the potential
We need to compute the partial derivatives of \( V \) with respect to \( x \) and \( y \).

1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x} \left( \frac{3x^2}{2} - \frac{y^2}{4} \right) = 3x \]

2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left( \frac{3x^2}{2} - \frac{y^2}{4} \right) = -\frac{y}{2} \]

### Step 3: Write the electric field vector
Using the results from the partial derivatives, we can express the electric field vector:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right) = -\left( 3x \hat{i} - \frac{y}{2} \hat{j} \right) \]
This simplifies to:
\[ \mathbf{E} = -3x \hat{i} + \frac{y}{2} \hat{j} \]

### Step 4: Substitute the point (1 m, 2 m)
Now, we substitute \( x = 1 \) m and \( y = 2 \) m into the electric field equation:
\[ \mathbf{E} = -3(1) \hat{i} + \frac{2}{2} \hat{j} = -3 \hat{i} + 1 \hat{j} \]

### Step 5: Write the final answer
Thus, the electric field intensity at the point (1 m, 2 m) is:
\[ \mathbf{E} = -3 \hat{i} + 1 \hat{j} \quad \text{(in N/C)} \]