1. 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. The capacitor has a voltage of 110v and is not connected to a battery. Calculate the capacitance. What is the new capacitance if we replace water with a new dielectric material with a constant of 3.75 in between the plates? What is the new voltage? What is the charge on each plate? 2. A capacitor of 3.23µF has an area of 6.35mm^2. Determine the separation distance between the two plates. If the new capacitance is now 5.63mF, what is the value of the dielectric inserted to the capacitor? What is the new voltage if the capacitor is connected to a 110 volts source? If the electric field created by two plates 8.99x10^4 N/C and a working voltage of 63 volts. Will the capacitor experience a dielectric breakdown? Prove your answer.

USE GRESA

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