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3 years ago

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The rectangular plates in a parallel-plate capacitor are 0.063 m × 5.4 m. A distance of 3.5 × 10–5 m separates the plates. The p

lates are separated by a dielectric made of Teflon, which has a dielectric constant of 2.1. What is the capacitance of the capacitor?

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

capacitance of the capacitor = 0.18 μ F

Explanation:

Area of the plate A = 0.063 m x 5.4 m = 0.3402 m²

distance between the plate d = 3.5 × 10–5 m

dielectric value for Teflon K = 2.1

capacitance of capacitor = ?

Formula for capacitance of parallel plate is as follows ,

$C= \frac{K\epsilon_0 A}{d}$ (Where $\epsilon_0 = 8.82 \times 10^-12\ \frac{F}{m}$ )

putting the values in the equation,

C = $\frac{2.1\times8.82\times10^-12\times0.3402 }{3.5\times10^-5}$ $=0.18\times 10^-6$ =0.18 μ F

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Answer:

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Explanation:

Given:

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position of second charge, $x_2=-1\ m$

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$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

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Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

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