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Bezzdna [24]
3 years ago
11

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field

magnitude E between them (neglect fringing) is 80 N/C. Find |q|.
Physics
1 answer:
Nuetrik [128]3 years ago
3 0

Answer:

q=6.22*10^-10C

Explanation:

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|

E=α/∈, electric field within the plate

α=q/A

A=area of the plate

∈=is the permittivity

substituting , we have

The field magnitude E between them (neglect fringing)

E=q/A∈

q=EA∈

q=0.88*80*8.84*10^-12

q=6.22*10^-10C

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Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

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1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

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Calculating:

\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}

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