PS508A: Solve Poisson’s equation to obtain expressions for the potential between a pair of large parallel plates separated by a distance L: Applied Behavior Analysis Assignment, NUI, Ireland
|University||National University of Ireland (NUI)|
|Subject||PS508A: Applied Behavior Analysis|
1: Solve Poisson’s equation to obtain expressions for the potential between a
pair of large parallel plates separated by a distance L and filled with a nonpolarisable neutral material for the following cases:
1.1 The two plates have the same potential, Φ0.
1.2 The two plates are at equal and opposite potentials (say, Φ = Φ0 at x = 0
and Φ = −Φ0 at x = L)
2: Consider the charge distributions shown below where the positive charges are uniformly distributed with charge density ni and the negative charge density, ne, is constant and equal to ni in the region x ∈ [s, L − s]. Take it that charge distribution is uniform and infinite in the ˆz and ˆy directions
2.1 Show that the potential in the region of electron depletion, x ∈ [0, s] is
where the potential of the walls containing the charge distribution is taken
as zero and Φs the potential at s +. You should assume that Φ′ (s) = 0. Plot Φ for x ∈ [0, s].
2.2 Use the result of 2.1 to obtain an expression for the the electric field for
x ∈ [0, s]. Plot Ex for x ∈ [0, s].
2.3 Justify the use of Φ′ (s) = 0. (HINT: use the symmetry of the charge
distribution and the result of question 1.1.)
Consider now a cylindrical case of the above distribution with ni = constant
and an electron density given by, where d is the radius of the neutral region and D the radius of the cylinder.
B.1 Derive an expression for the potential radially across a long cylinder using
analogous boundary condition as in Question 2.
B.2 Show that as D → ∞ the expression in B.1 goes to that in 2.1.
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