Physics, 04.02.2020 01:55 marly032205

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.

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Answer from: sihamabdalla591

A) C_3 < C_1 < C_2

The capacitance of a parallel-plate capacitor is given by

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

where

A is the plate area

d is the plate separation

Here we have:

- Capacitor 1: plate area A, plate separation d

capacitance: $C_1=\epsilon_0 \frac{A}{d}$

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

capacitance: $C_2=\epsilon_0 \frac{2A}{d} = 2C_1$

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

capacitance: $C_3=\epsilon_0 \frac{A}{2d}=\frac{C_1}{2}$

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

C_3 < C_1 < C_2

2. V_2 < V_1 < V_3

The three capacitors have same amont of charge, Q.

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

$V = \frac{Q}{C}$

so here we have

- Capacitor 1: C = C_1

Potential difference: $V_1 = \frac{Q}{C_1}$

- Capacitor 2: C = 2C_1

Potential difference: $V_2= \frac{Q}{2C_1}=\frac{ V_1}{2}$

- Capacitor 3: $C = \frac{C_1}{2}$

Potential difference: $V_3 = \frac{Q}{C_1/2}=2 V_1$

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

V_2 < V_1 < V_3

C. E_2 < E_1 = E_3

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

$E=\frac{V}{d}$

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

So here we have

- Capacitor 1: potential difference V_1 , plate separation d

electric field: $E_1 = \frac{V_1}{d}$

- Capacitor 2: potential difference $\frac{V_1}{2}$ , plate separation d

electric field: $E_2=\frac{V_1/2}{d} =\frac{V_1}{2d}= \frac{E_1}{2}$

- Capacitor 3: potential difference 2V_1 , plate separation 2d

electric field: $E_3=\frac{2 V_1}{2d} =\frac{V_1}{d}= E_1$

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

E_2 < E_1 = E_3

D. U_2 < U_1 < U_3

The energy stored in a capacitor is

$U=\frac{1}{2}QV$

where Q is the same for the three capacitors

Here we have

- Capacitor 1: potential difference V_1

energy: $U_1 = \frac{1}{2}QV_1$

- Capacitor 2: potential difference $\frac{V_1}{2}$

energy: $U_2 = \frac{1}{2}Q\frac{V_1}{2}=\frac{U_1}{2}$

- Capacitor 3: potential difference 2V_1

energy: $U_3 = \frac{1}{2}Q(2 V_1)=2 U_1$

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

U_2 < U_1 < U_3

E. u_2 < u_1 = u_3

The energy density in a capacitor is given by

$u=\frac{1}{2}\epsilon_0 E^2$

where E is the electric field strength

Here we have

- Capacitor 1: electric field E_1

Energy density: $u_1=\frac{1}{2}\epsilon_0 E_1^2$

- Capacitor 2: electric field $\frac{E_1}{2}$

energy density: $u_2=\frac{1}{2}\epsilon_0 (\frac{E_1}{2})^2=\frac{E_1}{4}$

- Capacitor 3: electric field E_1

Energy density: $u_3=\frac{1}{2}\epsilon_0 E_1^2$

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

u_2 < u_1 = u_3

Answer from: Quest

6 cos 34, 6 sin 34 trig

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