Two semi-infinite planar electrodes are arranged papally separated by a distance d. Between them there is a medium of conductivity g. If the potential difference between the electrodes is Φo, find: a) the current density vector at any point in the medium of conductivity g.b) the value of the surface charge density at any point of the armor connected to ground.

(a) Current Density Vector

To find the current density vector at any point in the medium, we can use Ohm’s law:

J = gE

where J is the current density vector, g is the conductivity of the medium, and E is the electric field.

The electric field between the two electrodes is uniform and directed from the positive electrode to the negative electrode. The magnitude of the electric field is given by:

E = Φo/d

Therefore, the current density vector at any point in the medium is given by:

J = gE = gΦo/d

The direction of the current density vector is parallel to the electric field, from the positive electrode to the negative electrode.

(b) Surface Charge Density

To find the surface charge density at any point on the electrode connected to ground, we can use the following equation:

σ = ε0E

where σ is the surface charge density, ε0 is the permittivity of free space, and E is the electric field at the surface of the electrode.

The electric field at the surface of the electrode is equal to the negative of the electric field in the medium, since the electrode is connected to ground. Therefore, the surface charge density at any point on the electrode connected to ground is given by:

σ = -ε0Φo/d

The surface charge density is negative, indicating that there is an excess of electrons on the surface of the electrode.

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