A

Step 1: The potential energy density u in an electric field is given by the formula:
$u = \frac{1}{2} \epsilon E^2$
where \( E \) is the electric field and \( \epsilon \) is the permittivity of free space.
Step 2: To find the electric field \( E \), we calculate the gradient of the potential function V:
$$ E = -\nabla V $$
where
$$ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) $$
Step 3: Calculate the partial derivatives:
$$ \frac{\partial V}{\partial x} = 2, \quad \frac{\partial V}{\partial y} = 4 $$
Thus, the electric field \( E \) can be expressed as:
$$ E = - \left( 2, 4 \right) = \left( -2, -4 \right) $$
Step 4: Calculate the magnitude of the electric field:
$$ |E| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} $$
Step 5: Assume permittivity of free space \( \epsilon \approx 8.854 \times 10^{-12} \text{ F/m} \):
$$ u = \frac{1}{2} \epsilon |E|^2 = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (20) \approx 8.854 \times 10^{-11} \text{ J/m}^3 $$
Step 6: Now calculate the energy stored in a 1-m^3 volume:
Total energy \( U \) in the volume is given by:
$$ U = u \times V = 8.854 \times 10^{-11} \times 1 = 8.854 \times 10^{-11} \text{ J} $$
Therefore, the stored energy in a 1-m^3 volume is approximately 8.854 x 10^{-11} J.