A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Paul has magnitude 43.0 N and direction 61.0∘ south of west.
How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
1 answer:
Answer:
W = -510.98J
Explanation:
Force = 43N, 61° SW
Displacement = 12m, 22° NE
Work done is given as:
W = F*d*cosA
where A = angle between force and displacement.
Angle between force and displacement, A = 61 + 90 + 22 = 172°
W = 43 * 12 * cos172
W = -510.98J
The negative sign shows that the work done is in the opposite direction of the force applied to it.
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HOPE THIS HELPS!!
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