The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates. Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?

a) required area is 1.1318 m²

b) the maximum potential difference that can be applied across the compactor is 1931.1 V

Explanation:

Given the data in the question;

dielectric constant εr = 2.35

distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m

dielectric strength = 49.5 MV/m

a)

given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F

To find the Area, we use the following the expression.

C = ε₀εrA / d

we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹²  (F/m)

we substitute

0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A  ] /  7.85 × 10⁻⁵

A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]

A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹

A = 1.1318 m²

Therefore, required area is 1.1318 m²

b)

the maximum potential difference that can be applied across the compactor.

We use the following expression;

⇒ 1/2 × dielectric strength × thickness d

we substitute

⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )

⇒ 1931.1 V

Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V

this means that the temperature of both substances will be equal

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: