Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;

Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
4a)the displacement is the distance moved in a direction but since no direction is given, the displacement is equal to the distance
b) the distance moved is 400m because that's the length of the track
Known :
l = 7 cm
w = 4 cm
Asked :
h = ...?
Answer :
V = B triangle × h (long)
35 = ½ × 4 × h × 7
35 = ½ × 28 × h
35 = 14 h
h = 35 ÷ 14
h = 2,5 cm
Sorry if I am wrong, I only study