Answer:
the result is the quantization of __Energy__ of the particle
Explanation:
What are the options? You did list them.
Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV
<span>I would say greater than because as you do deeper, the pressure strengthens. If you were in a 10 ft deep pool and you dive all the way to the bottom, the ears usually pop. That's because of the pressure. Whereas if you were to go five feet, your ears wouldn't. It depends on the age of the person.
Hope this helps.</span>
The negative charges of the balloon will stick to the positive charges on the wall.