A uniform electric fied exists in x-y plane. The potential of ponts `A(2m,2m),B(-2m,2m)` and `C(2m,3m)` are `4V, 16V` and `12V` respectively. The electric field is

Options

A. `(4hati+5hatj)V//m`
B. `(3hati+4hatj)V//m`
C. `-(3hati+4hatj)V//m`
D. `(3hati-4hatj)V//m`

Explanation

To find the electric field in the given scenario, we can use 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 components of the electric field can be calculated as:

\[ E_x = -\frac{dV}{dx}, \quad E_y = -\frac{dV}{dy} \]

### Step 1: Calculate \( E_x \)

We can find \( E_x \) by considering points A and B, where the y-coordinate remains constant. The coordinates are:
- \( A(2m, 2m) \) with \( V_A = 4V \)
- \( B(-2m, 2m) \) with \( V_B = 16V \)

Using the formula for \( E_x \):

\[ E_x = -\frac{V_A - V_B}{x_A - x_B} \]

Substituting the values:

\[ E_x = -\frac{4V - 16V}{2m - (-2m)} = -\frac{-12V}{4m} = \frac{12V}{4m} = 3 \, \text{V/m} \]

### Step 2: Calculate \( E_y \)

Next, we calculate \( E_y \) using points A and C, where the x-coordinate remains constant. The coordinates are:
- \( A(2m, 2m) \) with \( V_A = 4V \)
- \( C(2m, 3m) \) with \( V_C = 12V \)

Using the formula for \( E_y \):

\[ E_y = -\frac{V_C - V_A}{y_C - y_A} \]

Substituting the values:

\[ E_y = -\frac{12V - 4V}{3m - 2m} = -\frac{8V}{1m} = -8 \, \text{V/m} \]

### Step 3: Write the Electric Field Vector

Now that we have both components of the electric field, we can write the electric field vector \( \vec{E} \):

\[ \vec{E} = E_x \hat{i} + E_y \hat{j} = 3 \hat{i} - 8 \hat{j} \, \text{V/m} \]

### Final Answer

The electric field is:

\[ \vec{E} = 3 \hat{i} - 8 \hat{j} \, \text{V/m} \]

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