Question

HW.37. Redo Hw. 27 using the sump condition (0) and Gauss's law for Ē. As we did in HW.29, the symmetry in the problem leads to È (F) = Ez(z)? Although we did not use in solving HW. 27, there is a mirror symmetry across the XJ-plane. zo ok zo Suppose that there is an observer at z=2o facing the xy-plane.

Answer 1

a) From symmetry of the problem we have

$\vec{E} = E_{z}(z)\hat{z} \quad (1)$

and from mirror symmetry we have

$\vec{E}(x,y,-z) = -\vec{E}(x,y,z) \quad (2)$

Hence

Ēr, y, -0) = -Ēr, y, +0) (3)

Hence

$\vec{E}(x,y,+0).\hat{z} -\vec{E}(x,y,-0).\hat{z} = 2E_{z}(z=+0) = 4\pi\sigma\Rightarrow E_{z}(z=+0) = 2\pi\sigma$

Er, 4, +0) = 2πσά = E(.x, y, -0) =-2πσά (4)

b) Consider a cylinder with its axis oriented along the z axis and completely lying in the half of the space. Let the base of the cylinder be at $z=z_{1}$ and the top at $z=z_{2}$ , the flux of electric field through the surface of the cylinder is

$\iint_{S}\vec{E}.\vec{dS} = \iint_{A} E_{z}(z_{2})dxdy-\iint_{A} E_{z}(z_{1})dxdy = \iint_{A} \left(E_{z}(z_{2})-E_{z}(z_{1}) \right )dxdy \quad (5)$

Because of the form of electric field mentioned in equation (1) the dot product between the area vector and electric field is non-zero only at the base and top of the cylinder. Integral over signifies integral over the area of the top or bottom of the cylinder. Since this cylinder encloses no charge, integral form of Gauss law gives us

$\iint_{S}\vec{E}.\vec{dS} = 0\Rightarrow \left(E_{z}(z_{2})-E_{z}(z_{1}) \right )\iint_{A}dxdy = 0\Rightarrow E_{z}(z_{1}) = E_{z}(z_{2})$

$\Rightarrow \vec{E}(x,y,z_{2}) = \vec{E}(x,y,z_{1}) \quad (6)$

We have proved equation (6) for arbitrary positive numbers $z_{1} \text{ and } z_{2}$ , for arbitrary negative numbers similar construction and similar arguments can be used.

c) Using results of part (b) we get

$\vec{E}(x,y,z>0) = \vec{E}(x,y,+0) = 2\pi\sigma\hat{z} \quad (7)$

and

$\vec{E}(x,y,z<0) = \vec{E}(x,y,-0) = -2\pi\sigma\hat{z} \quad (8)$

Hence

$\vec{E}(\vec{r}) = 2\pi\sigma\hat{z},\; z>0 \quad (9)$

$\vec{E}(\vec{r}) = -2\pi\sigma\hat{z},\; z<0 \quad (10)$