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Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 2.0 cm from the center of the two surfaces?

Respuesta :

Answer:

E = K Q / R^2     by Gauss' Law where Q is the  charge enclosed by the surface of revolution and R is the distance from the enclosed charge

Since Q = d 4 pi r^2   where d is the charge density and r the radius of the inner sphere

E = K / R^2 * (4 d pi r^2) = 4 K pi d (r / R)^2 = 4 K pi d * 1/4

E =  9 * 10E9 * 3.14 * 40 * 10E-12 = 1.13 N / C

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