Question

A certain dielectric, with a dielectric constant 26, can withstand an electric field of 5 × 107 V/m. Suppose we want to use this dielectric to construct a 0.3 µF capacitor that can withstand a potential difference of 2000 V. The permittivity of free space is 8.85 × 10−12 C 2 /N · m2 . (a) What is the minimum plate separation? (b) What must the area of the plates be?

Answer 1

Answer: a) 40 µm b) 0.05 m2

Step-by-step explanation:

Assuming that the capacitor that we want to construct is of parallel plates type, we know that the Capacitance of this capacitor can be found using the following equation:

C= ε0 εr A / d

Now, we know that the dielectric, can withstand an electric field of 5 x 107 V/m, and between the plates of a capacitor, the electric field (which is constant for a given charge on the plates), and the voltaje between plates, are related by the following expression:

V = E. d, so, if the voltage between plates would be of 2000 V, and Emax can be of

5 x 107 V/m, we can find the mimimum d between plates, as follows:

dmin = V/ Emax = 2000 V / 5 x 107 V/m = 40 µm

Replacing this value in (1) above:

C= 8.85 × 10−12 C 2 /N · m2. 26. A / 40 µm

Solving for A, we have the following:

A = 0.3 µF.40 µm / 8.85 × 10−12 C 2 /N · m2. 26 = 0.05 m2

Step-by-step explanation: