Comments (9) on “User disconnected from your Channel”

A. Disconnected from. A participial phrase, as I like to think of it myself, is the phrase that has a word that looks like a verb but is an adjective or adverb. In this case, the participial phrase is an adjective and describes "we". Hope this helps.

Capacitance in a condensate can be found with any of the following equations

C = Q / DV

C = e or A / d

with these two expressions we answer the final statements

a) False. From the equations above we see that by decreasing the distance between the plates (d) the capacitance increases, by disconnecting the ideal capacitor the charge remains constant

b) False. After disconnecting the battery the charge on the plates remains constant

The energy stored in a capacitor is given by

U = k q / DV = k e A / d

c) True. From the previous equation we see that the energy is proportional to the dielectric

d) False. Capacitance increases with dielectric

e) False. The dielectric creates a field that opposes the field of the capacitor, whereby the total electric field decreases accordingly as the field and the voltage are proportional the potential difference must decrease

The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.

Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.

[tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex] [tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex] [tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex] [tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex]

A. Disconnected from. A participial phrase, as I like to think of it myself, is the phrase that has a word that looks like a verb but is an adjective or adverb. In this case, the participial phrase is an adjective and describes "we".

Hope this helps.

Its B

Explanation:

Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC

Explanation:

Since the same voltage in the battery is used for the entire rundown,

From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",

Q = C₁V = 30 μC

V = (30/C₁)

the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC

The charge on both capacitors are the same and equal to 15 μC (because they are in series)

Q = (Ceq) V = 15 μC

(Ceq) = (15/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (15/V) μF

(Ceq) = 15 ÷ (30/C₁)

(Ceq) = 15 × (C₁/30) = 0.5 C₁

(1/Ceq) = (2/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₂)

(2/C₁) = (1/C₁) + (1/C₂)

(2/C₁) - (1/C₁) = (1/C₂)

(1/C₁) = (1/C₂)

C₁ = C₂

C₂ = C₁

C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.

The charge on both capacitors are the same and equal to 10 μC (because they are in series)

Q = (Ceq) V = 10 μC

(Ceq) = (10/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (10/V) μF

(Ceq) = 10 ÷ (30/C₁)

(Ceq) = 10 × (C₁/30) = 0.333 C₁

(1/Ceq) = (3/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₃)

(3/C₁) = (1/C₁) + (1/C₃)

(3/C₁) - (1/C₁) = (1/C₃)

(2/C₁) = (1/C₃)

C₁ = 2C₃

C₃ = (C₁/2)

C₁, C₂, and C₃ are connected in series with one another and

with the battery, what is the charge on C₁

The charge on C₁ is the same as the charge on all the capacitors and equal to Q,

Q = (Ceq) V

(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)

Substituting for C₂ and C₃

C₂ = C₁ and C₃ = (C₁/2)

(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)

(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)

(1/Ceq) = (4/C₁)

Ceq = (C₁/4)

Q = (Ceq) V = (C₁/4) V

But recall that V = (30/C₁) from the first connection

Q = (C₁/4) (30/C₁)

Q = (30/4) = 7.5 μC

Hope this helps!

true C

false a, b, d, e

Explanation:

Capacitance in a condensate can be found with any of the following equations

C = Q / DV

C = e or A / d

with these two expressions we answer the final statements

a) False. From the equations above we see that by decreasing the distance between the plates (d) the capacitance increases, by disconnecting the ideal capacitor the charge remains constant

b) False. After disconnecting the battery the charge on the plates remains constant

The energy stored in a capacitor is given by

U = k q / DV = k e A / d

c) True. From the previous equation we see that the energy is proportional to the dielectric

d) False. Capacitance increases with dielectric

e) False. The dielectric creates a field that opposes the field of the capacitor, whereby the total electric field decreases accordingly as the field and the voltage are proportional the potential difference must decrease

yes sir

Explanation:

One would be A and two would be D and three would be C

Explanation:

The equation of the capacitance of the capacitor can be represented as:

[tex]C = \dfrac{\varepsilon_oA}{d}[/tex]

Also, the electric field between the plates can be expressed as:

[tex]E = \dfrac{\sigma}{\varepsilon_o}[/tex]

[tex]= \dfrac{Q}{A \varepsilon _o} \ \ (surface \ charge\ density \ \sigma =\dfrac{Q}{\varepsilon_o})[/tex]

However;

when pushed to a distance d/2, the new capacitance of the capacitor is:

[tex]C = \dfrac{\varepsilon _oA}{(d/2))}[/tex]

[tex]=\dfrac{2 \varepsilon_oA}{d}[/tex]

[tex]=2C \\ \\ Q' = CV \\ \\ = 2CV \\ \\ =2Q[/tex]

SImilarly, the new electric field between the plates is:

[tex]E' = \dfrac{\sigma'}{\varepsilon_o} \\ \\ = \dfrac{Q'}{A \varepsilon_o} \\ \\ = \dfrac{2Q}{A \varepsilon_o} \\ \\ =2E[/tex]

For Battery disconnected:

The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.

Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.

Hence, the ratio is :

[tex]\dfrac{E_{connected}}{E_{disconnected}} =\dfrac{2E}{E} \\ \\ \dfrac{E_{connected}}{E_{disconnected}} = 2[/tex]

Hence, the ratio is = 2

Mkay that’s pretty cool ig

The total charge is equal to 19.8meuC

Explanation:

For explanation see the picture attached

[tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex]

[tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex]

[tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex]

[tex]Consider three capacitors C1, C2, and C3 and a battery. If only C1 is connected to the battery, the[/tex]