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
It has 50kg with a velocity of 1 m/s times the speed of the cart divided by 2 and multiplied by kinectic x plus 5
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Bulbs c and b would still be screwed in if they were in to begin with and bulbs A, D, and E. would be unscrewed