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A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 9 ✕ 10−8 C at location −0.6, 0, 0 m. At location 0, 0.08, 0 m, what is the electric field contributed by the polarization charges on the surface of the metal sphere?

Respuesta :

Answer:

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

Explanation:

As we know that electric field inside any conducting shell is always zero

so we can say that

[tex]E_{charge} + E_{polarization} = 0[/tex]

here we know that

[tex]E_{charge} = \frac{kq}{r^2} \hat r[/tex]

here we know that

[tex]\hat r = \frac{(0 - -0.6)\hat i + (0.08 - 0)\hat j}{\sqrt{0.6^2 + 0.08^2}}[/tex]

[tex]\hat r = 0.99\hat i + 0.132\hat j[/tex]

now we will have

[tex]E_{polarization} = - E_{charge}[/tex]

[tex]E_{polarization} = - \frac{(9\times 10^9)(9 \times 10^{-8})}{0.6^2 + 0.08^2}(0.99\hat i + 0.132\hat j)[/tex]

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

okpalawalter8

We have that the electric field contributed by the polarization charges on the surface of the metal sphere is

  • E polarization=2.21*10^3(-0.99i-0.132j))

From the question we are told

  • A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 9 ✕ 10−8 C at location −0.6, 0, 0 m.
  • At location 0, 0.08, 0 m,
  • what is the electric field contributed by the polarization charges on the surface of the metal sphere?

E polarization

Generally the equation for the electric field inside any conducting shell    is mathematically given as

E charge+E polarization=0

Therefore

[tex]E charge=\frac{kq}{r^2}r'\\\\E charge=\frac{0--0.6i+0.08-0j}{\sqrt{0.6^2+0.08^2}}\\\\r'=0.99i+0.132j[/tex]

Therefore

E polarization=-E charge

Therefore

  • E polarization=2.21*10^3(-0.99i-0.132j))

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