contestada

Three parallel-plate capacitors each store the same amount of charge. Capacitor 1 has a plate area Aand a plate separation d. Capacitor 2 has a plate area 2A and a plate separation d. Capacitor 3 has a plate area A and a plate separation 2d.

Part A
Rank the three capacitors, based on their capacitance.
Part B
Rank the three capacitors, based on the potential difference between the plates.
Part C
Rank the three capacitors, based on the electric field magnitude between the plates.
Part D
Rank the three capacitors, based on the energy stored.
Part E
Rank the three capacitors, based on the energy density.

Respuesta :

skyluke89

A) [tex]C_3 < C_1 < C_2[/tex]

The capacitance of a parallel-plate capacitor is given by

[tex]C=\epsilon_0 \frac{A}{d}[/tex]

where

A is the plate area

d is the plate separation

Here we have:

- Capacitor 1: plate area A, plate separation d

capacitance: [tex]C_1=\epsilon_0 \frac{A}{d}[/tex]

- Capacitor 2: plate area 2A, plate separation d

capacitance: [tex]C_2=\epsilon_0 \frac{2A}{d} = 2C_1[/tex]

- Capacitor 3: plate area A, plate separation 2d

capacitance: [tex]C_3=\epsilon_0 \frac{A}{2d}=\frac{C_1}{2}[/tex]

So ranking the three capacitor from least to greatest capacitance we have:

[tex]C_3 < C_1 < C_2[/tex]

2. [tex]V_2 < V_1 < V_3[/tex]

The three capacitors have same amont of charge, Q.

The potential difference between the plates on each capacitor is given by

[tex] V = \frac{Q}{C}[/tex]

so here we have

- Capacitor 1: [tex]C = C_1[/tex]

Potential difference: [tex] V_1 = \frac{Q}{C_1}[/tex]

- Capacitor 2: [tex]C = 2C_1[/tex]

Potential difference: [tex] V_2= \frac{Q}{2C_1}=\frac{ V_1}{2}[/tex]

- Capacitor 3: [tex]C = \frac{C_1}{2}[/tex]

Potential difference: [tex] V_3 = \frac{Q}{C_1/2}=2 V_1 [/tex]

So ranking the three capacitor from least to greatest potential difference we have:

[tex]V_2 < V_1 < V_3[/tex]

C. [tex]E_2 < E_1 = E_3[/tex]

The electric field magnitude between the plates of a capacitor is given by

[tex]E=\frac{V}{d}[/tex]

where V is the potential difference between the plates and d is the plate separation

So here we have

- Capacitor 1: potential difference [tex]V_1[/tex], plate separation d

electric field: [tex]E_1 = \frac{V_1}{d}[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex], plate separation d

electric field: [tex]E_2=\frac{V_1/2}{d} =\frac{V_1}{2d}= \frac{E_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex], plate separation 2d

electric field: [tex]E_3=\frac{2 V_1}{2d} =\frac{V_1}{d}= E_1[/tex]

So ranking the three capacitor from least to greatest electric field we have:

[tex]E_2 < E_1 = E_3[/tex]

D. [tex]U_2 < U_1 < U_3[/tex]

The energy stored in a capacitor is

[tex]U=\frac{1}{2}QV[/tex]

where Q is the same for the three capacitors

Here we have

- Capacitor 1: potential difference [tex]V_1[/tex]

energy: [tex]U_1 = \frac{1}{2}QV_1[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex]

energy: [tex]U_2 = \frac{1}{2}Q\frac{V_1}{2}=\frac{U_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex]

energy: [tex]U_3 = \frac{1}{2}Q(2 V_1)=2 U_1[/tex]

So ranking the three capacitor from least to greatest energy we have:

[tex]U_2 < U_1 < U_3[/tex]

E. [tex]u_2 < u_1 = u_3[/tex]

The energy density in a capacitor is given by

[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]

where E is the electric field strength

Here we have

- Capacitor 1: electric field [tex]E_1[/tex]

Energy density: [tex]u_1=\frac{1}{2}\epsilon_0 E_1^2[/tex]

- Capacitor 2: electric field [tex]\frac{E_1}{2}[/tex]

energy density: [tex]u_2=\frac{1}{2}\epsilon_0 (\frac{E_1}{2})^2=\frac{E_1}{4}[/tex]

- Capacitor 3: electric field [tex]E_1[/tex]

Energy density: [tex]u_3=\frac{1}{2}\epsilon_0 E_1^2[/tex]

So ranking the three capacitor from least to greatest energy density we have:

[tex]u_2 < u_1 = u_3[/tex]

Otras preguntas