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olga55 [171]
3 years ago
8

Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge

on each plate in nanoCoulombs.
Physics
1 answer:
Arisa [49]3 years ago
3 0

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

   =[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

   = 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

   = (7.86×10^-14)×(61)

   = 4.80×10^-14 C

   ≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

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Answer:

12.245m3

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Electric energy is calculated as

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Note that this energy has been given and is 60Joules

From conservation of energy it means;

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From change of subject of formula for M; we have:

M = 60/ g × h

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= 12.245kg

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Note the density of water is 1kg/m3

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Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s
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