Answer:
Explanation:
Electric field E = 4 x 10⁷ V / m
Dielectric constant k = 24
capacitance of capacitor
C = kε₀ A / d
d = plate separation
A = plate area
C = .89 x 10⁻⁶
V / d = electric field
for minimum d , electric field will be maximum
V / d = 4 x 10⁷
1930 / d = 4 x 10⁷
d = 1930 / 4 x 10⁷
d = 482.5 x 10⁻⁷ m
= 48.25 x 10⁻⁶ m
C = kε₀ A / d
.89 x 10⁻⁶ = 24 ε₀ A / d
A = .89 x 10⁻⁶ X d / 24 ε₀
A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²
= 42.9 / 212.4
= .2019 m²
Answer:
I think its B but I may be wrong
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:

Explanation:
From the question we are told that:
Electric field 
Distance 
At negative plate
Generally the equation for Velocity is mathematically given by

Therefore




Answer:
Light's angle of refraction = 37.1° (Approx.)
Explanation:
Given:
Index of refraction = 1.02
Base of refraction = 1
Angle of incidence = 38°
Find:
Light's angle of refraction
Computation:
Using Snell's law;
Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction
Sin38 / Light's angle of refraction = 1.02 / 1
Sin[Light's angle of refraction] = Sin 38 / 1.02
Sin[Light's angle of refraction] = [0.6156] / 1.02
Sin[Light's angle of refraction] = 0.6035
Light's angle of refraction = 37.1° (Approx.)
I think god did ??? I searched it up okay