Question

A "homemade" capacitor can be built by placing a sheet of paper between two foil. Imagine that the sheet of paper is 21 cm x 30 cm and a thickness of 0.10 mm. a) What would be the capacitor capacity? b) What charge does it store if it is connected to a 12V battery? c) What maximum voltage could it support?

Answer 1

a) The capacitance is 12.8 nF

b) The charge stored is 154 nC

c) The maximum voltage supported is 1600 V

Explanation:

a)

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k \epsilon_0 A}{d}$

where

k is the relative permittivity of the material between the plates

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of each plate of the capacitor

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.3 for paper

$d = 0.10 mm = 1\cdot 10^{-4}m$

$A=(21 cm)(30 cm) = (0.21 m)(0.30 m)=0.063 m^2$

Therefore, the capacitance is

$C=\frac{(2.3)(8.85\cdot 10^{-12})(0.063)}{1\cdot 10^{-4}}=1.28\cdot 10^{-8} F = 12.8 nF$

b)

The charge stored on a capacitor is related to the potential difference applied by the equation:

Q = CV

where

C is the capacitance

V is the potential difference applied

In this problem, we have

$C=1.28\cdot 10^{-9}F$

V = 12 V

Substituting,

$Q=(1.28\cdot 10^{-8})(12)=1.54\cdot 10^{-7} C = 154 nC$

c)

The dielectric strength for paper is

$E=16 MV/m = 16\cdot 10^6 V/m$

Which means that this is the maximum electric field that can be supported in the capacitor before charges start to flow from one plate to the other, crossing the gap.

In a parallel plate capacitor, the electric field is uniform, so it is related to the potential difference by

$V=Ed$

where d is the separation between the plates.

Therefore, by substituting:

$E=16\cdot 10^6 V/m$

and

$d=1\cdot 10^{-4} m$

We find the maximum voltage supported by the capacitor:

$V=(16\cdot 10^{6})(1\cdot 10^{-4})=1600 V$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly