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Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potentialdifferencebetweenthemis360V.(a)Whatisthemagnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge

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adelekeademolu

Answer:

a. 8000 N/C or 8000 V/m

b. Incomplete question

Explanation:

a. The electric field, E, is the ratio of the potential difference,V, to the separation distance, d.

V = 360 V

d = 45 mm = 0.045 m

[tex]E = \dfrac{V}{d} = \dfrac{360}{0.045} = 8000[/tex]

b. The force on a charge in an electric field is the product of the field and the charge.

[tex]F = qE[/tex]

With the charge, q, known, the force can be calculated.

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