1. For a parallel plate capacitor, the energy stored is

$U = \frac{1}{2}CV^2$

2. For a parallel plate capacitor,

$C = \frac{\varepsilon_0 A}{d}$

and the potential difference between the plates is

$V = Ed$

3. Substitute these into the energy formula:

$U = \frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(Ed)^2$

4. Simplify:

$U = \frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)E^2 d^2$

$U = \frac{1}{2}\varepsilon_0 E^2 Ad$

5. Hence, the potential energy stored in the capacitor is

$\boxed{\frac{1}{2}\varepsilon_0 E^2 Ad}$

6. Checking options:

• A: $\varepsilon_0 E^2 Ad$ ❌
• B: $\frac{3}{4}\varepsilon_0 E^2 Ad$ ❌
• C: $\frac{1}{4}\varepsilon_0 E^2 Ad$ ❌
• D: $\frac{1}{2}\varepsilon_0 E^2 Ad$ ✅

Therefore, the correct answer is D.