Question

S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initially uniform electric field →E = E₀ k^, the resulting electric potential is V(x, y, z)=V₀ for points inside the sphere andV(x, y, z)=V₀ - E₀z + E₀a³z / (x² + y² + z² )³/²for points outside the sphere, where V₀ is the (constant) electric potential on the conductor. Use this equation to determine the x, y , and z components of the resulting electric field (a) inside the sphere.

Answer 1

The sphere has a constant potential. It is the electric field.

$E = V_{0} = 0$

In the sphere, then

$E_{x} = 0, E_{y}=0, E_{z}=0$

Outside the sphere, then

$V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2} }$

The elements of the electric field include

$E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}$

Which becomes,

$=E_{0} (1-\frac{a^{3}}{x^{2} +y^{2}+z^{2})^{3/2}}+\frac{3a^{3}z^{2}}{(x^{2} +y^{2}+z^{2})^{5/2}})$

In a consistent electric field, is force constant?

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

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