Question 6.2: The xerographic copying machine is an important application ...

Consider the modeled version of Figure 6.2(a) shown in Figure 6.2(b). Since \rho_{v}=0 in this case, we apply Laplace’s equation. Also the potential depends only on x. Thus

\nabla^2 V=\frac{d^{2}V}{dx^{2}}=0

Integrating twice gives

V=Ax+B

Let the potentials above and below x=a be V_{1} and V_{2}, respectively:

V_{1}=A_{1}x+B_{1},       x\gt a                                                                      (6.2.1a)

V_{2}=A_{2}x+B_{2},       x\lt a                                                                     (6.2.1b)

The boundary conditions at the grounded electrodes are

V_{1}(x=d)=0                                                                            (6.2.2a)

V_{2}(x=0)=0                                                                            (6.2.2b)

At the surface of the photoconductor

V_{1}(x=a)=V_{2}(x=a)                                                                               (6.2.3a)

D_{1n}-D_{2n}=\rho_{S}|_{x=a}                                                                                (6.2.3b)

We use the four conditions in eqs. (6.2.2) and (6.2.3) to determine the four unknown constants A_{1},  A_{2},  B_{1}, and B_{2}. From eqs. (6.2.1) and (6.2.2),

0=A_{1}d+B_{1}\rightarrow B_{1}=-A_{1}d                                                                     (6.2.4a)

0=0+B_{2}\rightarrow B_{2}=0                                                                              (6.2.4b)

From eqs. (6.2.1) and (6.2.3a)

A_{1}a+B_{1}=A_{2}a                                                                               (6.2.5)

To apply eq. (6.2.3b), recall that D=\varepsilon E=-\varepsilon\nabla V so that

\rho_{S}=D_{1n}-D_{2n}=\varepsilon_{1}E_{1n}-\varepsilon_{2}E_{2n}=-\varepsilon_{1}\frac{dV_{1}}{dx}+\varepsilon_{2}\frac{dV_{2}}{dx}

or

\rho_{S}=-\varepsilon _{1}A_{1}+\varepsilon_{2}A_{2}                                                                     (6.2.6)

Solving for A_{1} and A_{2} in eqs. (6.2.4) to (6.2.6), we obtain

E_{1}=-A_{1}a_{x}=\frac{\rho_{S}a_{x}}{\varepsilon _{1}\left[1+\frac{\varepsilon _{2}}{\varepsilon _{1}}\frac{d}{a}-\frac{\varepsilon _{2}}{\varepsilon_{1}} \right]},       a\leq x\leq d

E_{2}=-A_{2}a_{x}=\frac{-\rho_{S}\left(\frac{d}{a}-1\right)a_{x}}{\varepsilon _{1}\left[1+\frac{\varepsilon _{2}}{\varepsilon _{1}}\frac{d}{a}-\frac{\varepsilon _{2}}{\varepsilon_{1}} \right]},       0\leq x\leq a