Answer:
Explanation:
The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:
Where A = Area of the plate
d = distance between the plates
dielectric constant
Case 1:
When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a
Therefore the capacitance of the capacitor becomes:
.......................(1)
Case 2:
Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:
.........................(2)
Equating (1) and (2)
Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W = 
.................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W = 
</em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.
Answer:
The charge flows in coulombs is
Explanation:
The current magnitude of current is given by the resistance and the induced Emf as:
, 
, 
, 
Ω
,
Replacing :
Answer:
a
Explanation:
because it has more energy
you can check attachment for answer.
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