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3 years ago

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Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

lower plate, and they are separated by 3 cm. You move the two plates carefully to a separation of 4 cm. What is the strength of the electric field between the plates?

1 answer:

BlackZzzverrR [31]3 years ago

5 0

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

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Answer:

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(b) L = 182.56 m

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Explanation:

The resistance of the wire is given as:

R = ρL/A

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(a)

For Gold Wire:

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(b)

For Copper Wire:

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1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

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<u>L = 182.56 m</u>

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(c)

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