Question

2. Imagine a parallel-plate capacitor with a plate separation of 1 cm. The plates are both right triangles of 2 cm on two sides. What is
the capacitance of this capacitor?​

Answer 1

Answer:

1.77 * 10^(-13)  F

Explanation:

C = (epsolen not *A)/d

= (8.85*10^-12  * ( 1/2 *(2*10^-2)^2 ) )/ (1* 10^-2)

=  1.77 * 10^(-13)

#LincolnFam

Answer 2

The capacitance value of the given parallel-plate capacitor is   $1.77*10^-^1^1 C^2/N.m$

Given:

A parallel-plate capacitor with plate separation: 1 cm

The given plates are right-angled triangles with a side length of 2 cm

To Find:

The capacitance of this capacitor

Solution:

The formula for the capacitance of a parallel-plate capacitor is $C = \frac{E_0A}{D}$

Where $E_0$ is the permeability of free space with a value $8.85$×$10^-^1^2$$C^2/N.m^2$

D is the distance between the plates, which is already given as 1 cm.

A is the are of the plate

⇒ A = $\frac{1}{2} (base)(height)$ = $\frac{1}{2}(2)(2)$

⇒ A = $2 cm^2$

⇒ $C = \frac{(8.85*10^-^1^2)(2)}{1}$

⇒ C ≈ $1.77*10^-^1^1 C^2/N.m$

∴ The capacitance value of the given parallel-plate capacitor is   $1.77*10^-^1^1 C^2/N.m$