The electric potential V at any point (x, y) in a plane is given by `V=5x^(2)-4y^(2)` volt. Find the magnitude of electric field intensity at the point (1m, 2m).

Explanation

To find the magnitude of electric field intensity at the point (1m, 2m) given the electric potential \( V = 5x^2 - 4y^2 \), we will follow these steps:

### Step 1: Understand the Relationship Between Electric Field and Electric Potential
The electric field intensity \( \vec{E} \) is related to the electric potential \( V \) by the negative gradient of the potential:
\[ \vec{E} = -\nabla V \]
In two dimensions, this can be expressed as:
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