Question

Potential \( V = \frac{1}{2}(y^2 - 4x) \). Electric field at \( x = 1 \, \text{m}, y = 1 \, \text{m} \) is:

• A. \( 2\hat{i} + \hat{j} \, \text{Vm}^{-1} \)
• B. \( -2\hat{i} + \hat{j} \, \text{Vm}^{-1} \)
• C. \( 2\hat{i} - \hat{j} \, \text{Vm}^{-1} \)
• D. \( -2\hat{i} + 2\hat{j} \, \text{Vm}^{-1} \)

Hint:

Electric field is negative gradient of potential: \( \vec{E} = -\nabla V \)

Answer 1

The Correct Option is C

\[ V = \frac{1}{2}(y^2 - 4x) \Rightarrow E_x = -\frac{\partial V}{\partial x} = 2, \quad E_y = -\frac{\partial V}{\partial y} = -y = -1 \Rightarrow \vec{E} = 2\hat{i} - \hat{j} \, \text{Vm}^{-1} \]