Question

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?

Answer 1

Answer:

$E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)$

Explanation:

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

so we can say that

$E_{charge} + E_{polarization} = 0$

here we know that

$E_{charge} = \frac{kq}{r^2} \hat r$

here we know that

$\hat r = \frac{(0 - -0.6)\hat i + (0.08 - 0)\hat j}{\sqrt{0.6^2 + 0.08^2}}$

$\hat r = 0.99\hat i + 0.132\hat j$

now we will have

$E_{polarization} = - E_{charge}$

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

$E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)$