Question

70. When an uncharged conducting sphere of radius a is placed at the origin of an xyz coordinate system that lies in an initially uniform electric field E = E, Î, the resulting electric potential is V(x, y, z) = V, for points inside the sphere and E, az V(x, y, z) = V. – Eoz + (x2 + y2 + z2)3/2 for points outside the sphere, where V, is the (constant) electric potential on the conductor. Use this equation to determine the x, y, and z components of the result- ing electric field (a) inside the sphere and (b) outside the sphere.

If you can draw a FBD as well

Answer 1

The potential inside the sphere is constant. The electric field is

$\vec E=-\vec \nabla V_0=0$

Hence inside the sphere

$E_x=0,\,\,\,E_y=0,\,\,\,E_z=0$

b) Outside the sphere, the field is

$V=V_0-E_0 z+\frac{E_0 a^3 z}{\left(x^2+y^2+z^2\right)^{3/2}}$

The electric field components are

$E_x=-\frac{\partial V}{\partial x}=\frac{3 E_0\,a^3 \, x\, z}{\left(x^2+y^2+z^2\right)^{5/2}}$

$E_y=-\frac{\partial V}{\partial y}=\frac{3 E_0\,a^3 \, y\, z}{\left(x^2+y^2+z^2\right)^{5/2}}$

and

$E_z=-\frac{\partial V}{\partial z}=E_0\left(1-\frac{a^3 }{\left(x^2+y^2+z^2\right)^{3/2}}+\frac{3 a^3 \, z^2}{\left(x^2+y^2+z^2\right)^{5/2}}\right)$