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Troyanec [42]
3 years ago
9

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

mm^2. How much charge must be transferred from one plate to the other if 1.1 nJ of energy are to be stored in the plates? ( ε0 = 8.85 × 10^-12 C2/N · m^2)
Physics
1 answer:
julia-pushkina [17]3 years ago
8 0

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

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7 0
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andrew-mc [135]

Answer:

high, low

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A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h
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If the vertical component is 29.6 m/s down, and the horizontal component
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