Suppose that we have a 1000 pF parallel-plate capacitor with air dielectric charged to 1000 V. The capacitors terminals are open
circuited. Find the stored energy. If the plates are moved apart so that d is doubled, determine the new voltage on the capacitor and the new stored energy. Where did the extra energy come from?
1 answer:
Answer:
A) initial stored energy = 5x10^-4 J
B) new voltage if space is doubled will be 2000V
C) if space is doubled, stored energy becomes 2x10^-3 J
D) the extra energy comes from the work done in moving the plates further apart
Explanation:
Detailed explanation and calculation is shown in the image below
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Answer:
γ =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN