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Mashutka [201]
3 years ago
12

Pls help

Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

sorry I dont now the answer bro i am so sorry xd ;'(

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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

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a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

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= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

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E_2 = 16.76 .

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