Answer:
F= σ² L² /2ε₀
F = (L² ε₀/4π) ΔV² / r⁴
Explanation:
a) For this exercise we can use Coulomb's law
F = - k Q² / r²
where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates
Capacitance is defined by
C = Q / ΔV
Q = C ΔV
also the capacitance for a parallel plate capacitor is related to its shape
C = ε₀ A / r
we substitute
Q = ε₀ A ΔV / r
we substitute in the force equation
F = k (ε₀ A ΔV / r)² / r²
k = 1 / 4πε₀
F = ε₀ /4π L² ΔV² / r⁴4
F = L² ΔV² ε₀/ (4π r⁴)
F = (L² ε₀/4π) ΔV² / r⁴
b) Another way to solve the exercise is to use the relationship between the force and the electric field
F = q E
where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface
Ф = ∫E .dA =
/ ε₀
the plate have two side
2E A = q_{int} / ε₀
E = σ / 2ε₀
σ = q_{int} / A
substituting in force
F = q σ / 2ε₀
the charge total on the other plate is
q = σ A
q = σ L²
F= σ² L² /2ε₀
13200N
Explanation:
Given parameters:
Mass = 1100kg
Velocity = 24m/s
time = 2s
unknown:
Braking force = ?
Solution:
The braking force is the force needed to stop the car from moving.
Force = ma = 
m is the mass of the car
v is the velocity
t is the time taken
Force =
= 13200N
Learn more:
Force brainly.com/question/4033012
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Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))