# Charge plate in an ionic solution

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The situation is as follows: Put a charged plate in a solution - the sign of the charge doesn't matter - in response, counterions will form in the solution such that - at infinite - the total electric field cancels. However, nearer the plate, you have a distribution of these counterions that is determined by the potential - V - due to the plate and the counterions. You only have to worry about one dimension - the distance from the plate in the x direction. Solve for V and rho where rho (r) is charge density The formula for the potential is as follows (Sorry for writing it out like this, I don't have word and the posting thing turns Mathematica symbols to gibberish):

d^2V/dx^2 = (1/ee(0))*r(0)*EXP(-z*e*V/kT)

My 2 boundary conditions are:

1.The electric field E= -dV/dx here - is zero at infinity

2. The potential V at the surface of the plate is zero.

After this, I need to find the charge distribution using the formula rho(r) = r(0)*EXP(-z*e*V/kT)

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##### Solution Summary

A detailed explanation is given. It is explained that the boundary condition at infinity cannot be satisfied. However, one can replace that boundary condition by another boundary condition and solve the problem that way.

##### Solution Preview

The differential equation is not difficult to solve. However, it is not possible to impose the boundary condition that the electric field is zero at infinity in the solution. This is caused by the fact that the assumption that the charge distribution is given by:

r(0)*EXP(-z*e*V/kT)

has to break down if you move away from the plate. This is because there the effect of the opposite charged ions is important.

To solve the differential equation you can proceed as follows. The diff. equation is:

d^2V/dx^2 = rho(0)/epsilon_{0}Exp[-z e V/(k T)]

Multiply both sides with the first derivative of V. The left side becomes:

d^2V/dx^2 dV/dx. This term can be integrated. After integration it is:

1/2 (dV/dx)^2.

The right hand side can now also ...

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