Electric Potential: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

The electric field is given by:
\[
\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right).
\]

Given \( V = \frac{1}{2}(y^2 - 4x) \):

1. Partial derivatives:
- \(\frac{\partial V}{\partial x} = -2\)
- \(\frac{\partial V}{\partial y} = y\).

2. At \( x = 1, y = 1 \):
- \( E_x = -\frac{\partial V}{\partial x} = -(-2) = 2 \),
- \( E_y = -\frac{\partial V}{\partial y} = -(1) = -1 \).

3. Electric field:
\[
\vec{E} = 2\hat{i} - \hat{j} \, \text{V/m}.
\]