Question

Given a 9.7 pF air-filled capacitor, you are asked to convert it to a capacitor that can store up to 7.7 µJ with a maximum potential difference of 550 V. What must be the dielectric constant of the material that you should you use to fill the gap in the air capacitor if you do not allow for a margin of error?

Answer 1

Answer:

ε = 5.24

Explanation:

In order to find the values of the dielectric constant of some material, which allows one to store an energy  of 7.7µJ, you use the following formula:

$E=\frac{1}{2}CV^2$       (1)

C: capacitance  = 9.7 pF

V: potential difference on the capacitor = 550 V

E: energy storage on the capacitor

Furthermore, you take into account that, when a dielectric material is included in the capacitor, its capacitance is modified as follow:

$C'=\epsilon C$

To find the value of ε, you use the equation (1), but for C = C'.

$E=\frac{1}{2}C'V^2=\frac{1}{2}\epsilon CV^2$      (2)

Then, you solve for ε in the previous equation:

$\epsilon=\frac{2E}{CV^2}=\frac{2(7.7*10^{-6}J)}{(9.7*10^{-12}F)(550V)^2}\\\\\epsilon=5.24$

The dielectric constant of the material required is 5.24

Answer 2

The dielectric constant of the material that you should you use to fill the gap in the air capacitor should be 5.24

Calculation of the dielectric constant of the material:

Since Given a 9.7 pF air-filled capacitor, you are asked to convert it to a capacitor that can store up to 7.7 µJ with a maximum potential difference of 550 V.

E = 1/2 CV^2

E = 2\CV^2

= 2(7.7*10^-6J)/ (9.7*10^-12F) (550 V)^2

= 5.24

Hence, The dielectric constant of the material required is 5.24

Learn more about air here: https://brainly.com/question/18820741