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A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 9.5 cm from the disk?

Respuesta :

Hashirriaz830

Answer:

[tex]=6.3*10^3 N/C[/tex]

Explanation:

solution:

from this below equation (1)

[tex]E=[/tex]σ/2εo[tex](1-\frac{z}{\sqrt{z^2-R^2} } )[/tex]...........(1)

we obtain:

[tex]=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )[/tex]

[tex]=6.3*10^3 N/C[/tex]

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