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3 years ago

A capacitor is connected to a power source and begins charging.

Physics

2 answers:

Drupady [299]3 years ago

5 0

Answer:

Electrons are too strongly repelled from the negatively charged plate.

Explanation:

A capacitor is a device that stores energy by holding apart pairs of opposite charges. The opposite charges are put in parallel plates. When it's charged by a power source, it's source is sending electrons to the capacitor, thus, it will be put in the negatively charged plate. At some point, the attractive force between the plates will be weaker than the repulsive forces of the electrons and the negatively charged plate, then, they'll be repelled and the capacitor will stop charging.

pashok25 [27]3 years ago

3 0

D.Electrons are too strongly repelled from the negatively charged plate.

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

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${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

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${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

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☯ As we know that,

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

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$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

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$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

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$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

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$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

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$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

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$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

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$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

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$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

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$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

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