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Rashid [163]
3 years ago
5

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0

V battery.
Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

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Answer:

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Given:

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\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

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\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

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\sf \implies s =  {10}^{2}

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Answer:

Explanation:

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