Question

The voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 10⁶ V/m. The plate area is 10^-4 m². What is the dielectric constant if the capacitance is 15 pF?

(given ε₀ = 8.86 × 10^-12 C²/Nm²)

पर्याय

• 3.8

• 8.5

• 4.5

• 6.2

Answer 1

8.5

Explanation:

Given: A parallel plate capacitor has a voltage rating of V = 500V, a dielectric strength of E = 10⁶ V/m, an area of the capacitor plate of A = 10^-4 m², and a capacitance of C = 15 pF, ε₀ = 8.86 × 10^-12 C²/Nm².

To find: K, The dielectric constant of the dielectric material used between the plates of the given capacitor.

Parallel plate capacitor capacitance:

`C = (ε_0KA)/d`

K = `(Cd)/(ε_0A)` ........(i)

In equation (i) d represents the distance between the capacitor's plates.

About the applied voltage and dielectric power:

`d = V/E` ......(ii)

Substitute the following from equation (ii) in equation (i):

K = `C/(ε_0A)(V/E)` .................(iii)

K = `(15 xx 10^-12)/(8.86 xx 10^-12 xx 10^-4) xx (500/10^6) ≈ 8.5`