Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Answer:
The force is 3305.6 N.
Explanation:
Final velocity, v = 0
time, t = 2.25 ms
initial velocity, u = 4.25 m/s
mass, m = 1.75 kg
Let the acceleration is a.
Use first equation of motion.
v = u + a t
0 = 4.25 + a x 0.00225
a = - 1888.9 m/s^2
The force is
F = ma
F = 1.75 x 1888.9
F = 3305.6 N
I believe the answer would be oxygeon deficit.
Answer:
r = 2.56 10⁻¹² m
Explanation:
For this exercise let's use energy conservation
Starting point, proton too far
Em₀ = K = ½ m v²
Final point. When proton is stop
Emf = U = k q₁ q₂ / r
How energy is conserved
Em₀ = Emf
½ m v² = k q₁ q₂ / r
r = 2k q₁ q₂ / m v²
Let's calculate
r = 2 8.99 10⁸ 1.6 10⁻¹⁹ 1.82 10⁻¹⁸ / (1.67 10⁻²⁷ (3.5 10⁵)² )
r = 2.56 10⁻¹² m