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A point charge of −3.50 µc is located in the center of a spherical cavity of radius 5.00 cm inside an insulating charged solid. the charge density in the solid is ρ = 7.35 × x10−4 c/m3 . calculate the electric field inside the solid at a distance of 10 cm from the center of the cavity.

Respuesta :

As per Gauss law

[tex]\int E. dA = \frac{q}{epsilon_0}[/tex]

here

q = total charge inside the gaussian surface

now in order to find q

[tex]q = q_1 + \rho* \frac{4}{3}\pi(R^3 - a^3)[/tex]

[tex]q = -3.50 \mu C + 7.35 * 10^{-4}* \frac{4}{3}\pi(0.10^3 - 0.05^3)[/tex]

[tex]q = -2.86\mu C[/tex]

now electric field is given by

[tex]E* 4\pi *0.10^2 = \frac{-2.86 \muC}{\epsilon_0}[/tex]

[tex]E = \frac{9*10^9*2.86*10^{-6}}{0.10^2}[/tex]

[tex]E = 2.57 * 10^6 N/C[/tex]

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