Answer:
A)
i) capacitance = 5.4 8 10^-11 f
ii) new capacitance when dielectric is change to material = 2.03 * 10^-11 f
iii) new voltage = 2926 v
iv) charge on each plate = 5.94 * 10^-9 c
B)
i) separation between plates = 1.74 * 10^11 m
ii) value of dielectric = 1743
iii) new voltage = 1.56*10^-6 v
iv ) the capacitor will not experience a dielectric breakdown because its working voltage : 63 v ∠ 110V
Explanation:
Attached below is a detailed solution to the given problems above
A) A capacitor is made of 2 rectangular metal plates with side length of 3cmx6cm separated by a distance of 2.36cm with water in between the plates.
i) capacitance = 5.4 8 10^-11 f
ii) new capacitance when dielectric is change to material = 2.03 * 10^-11 f
iii) new voltage = 2926 v
iv) charge on each plate = 5.94 * 10^-9 c
B) A capacitor of 3.23µF has an area of 6.35mm^2.
i) separation between plates = 1.74 * 10^11 m
ii) value of dielectric = 1743
iii) new voltage = 1.56*10^-6 v
iv ) the capacitor will not experience a dielectric breakdown because its working voltage : 63 v ∠ 110V
Answer:
I can only answer one question. I suggest you split the questions and upload them separately.
2. Yes the capacitor will experience dielectric breakdown with a new voltage of 1.56x10^-6v.
Explanation:
Dielectric breakdown of a capacitor occurs when the capacitor loses its conductivity due to the installation of high insulating materials.
The formula for capacitance is C = ∈οΑ/d
where ∈ο = 8.85x10^-12
A = 6.35x10^-6mm^2
C = 3.23μF
d = 8.85x10^-12 x 6.35x10^-6 / 3.23x10-6
The distance between the plate d = 1.74x10^-11 m
The constant between the old and new capacitance = 5.63mF
is C1 = KC
K = C1/C = 5.63x10^-3/ 3.23x10^-6 = 1743
When the capacitor is connected to a voltage of 110v;
v = 110 / 1743 = 0.063v
the new voltage V = ED
= 8.99x10^4 x 1.74x10^-11 = 1.56x10^-6v
