Chapter 27 - Gauss's Law - Exercises and Problems - Page 808: 50

ee the detailed answer below.

$$\color{blue}{\bf [a]}$$ The net electric field inside the conductor is zero, so the net electric fields at points 2 and 4 are zeros. $$\boxed{ E_{2}= E_{4}=\bf 0 \;\rm N/C}$$ To find the electric field at the other 3 points which are 1, 3, and 5, we need to use Gaussian surface so we drew a long cylinder that pass over the two slabs. According to Gauss's law, $$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$ $$\int_{\rm top} \vec Ed\vec A+\int_{\rm bottom} \vec Ed\vec A+\int_{\rm sides} \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$ As we see in the figure below, since the two slabs are positively charged, the electric field at point 1 must be upward while the electric field at point 5 must be downward. The electric field is zero thought the gaussian cylinder 1, so that $$E_1A_{\rm top}+E_5A_{\rm bottom}+0=\dfrac{Q_{in}}{\epsilon_0}$$ where $A_{\rm top}=A_{\rm bottom}=A'$ $$E_1A'+E_5A'=\dfrac{Q_{in}}{\epsilon_0}\tag 1$$ Now we need to find the charge enclosed inside our gaussian cylinder where we know that the charge area density is constant for the two slabs. The enclosed charge from the upper slab is given by $$\dfrac{Q}{A}=\dfrac{Q_{\rm in,upper}}{A'}$$ $$Q_{\rm in,upper}=\dfrac{QA'}{A}\tag 2$$ The enclosed charge from the lower slab is given by $$\dfrac{2Q}{A}=\dfrac{Q_{\rm in,lower}}{A'}$$ $$Q_{\rm in,lower}=\dfrac{2QA'}{A}\tag 3$$ Hence, $$Q_{in}=Q_{\rm in,upper}+Q_{\rm in,lower}$$ Plug from (2) and (3), $$Q_{in}=\dfrac{QA'}{A}+\dfrac{2QA'}{A}$$ $$Q_{in}= \dfrac{3QA'}{A}\tag 4$$ Plug from (4) into (1), $$E_1A'+E_5A'=\dfrac{ 3QA'}{A\epsilon_0}$$ $$E_1 +E_5 =\dfrac{ 3Q }{A\epsilon_0}$$ We can see that $E_1=E_5$ since we know that the field of a plane of charge is constant regardless of the distance from the plane and, in our case, the superposition of the field at points above the top plane should have the same magnitude but opposite direction as the superposition at points below the bottom plane. Hence, $$\boxed{E_1=E_5=\dfrac{ 3Q }{2\epsilon_0A}}$$ But $\vec E_1$ is pointing upward while $\vec E_5$ is pointing downward. Now we need to find the electric field at region 3, so we drew a new gaussian surface that passes through the upper slab only. It extracts from region 1 to region 3. Since the net charge of the lower slab is greater than that of the upper one, the net electric field at region 3 must be upward, as we see in the figure below. By the same approach, $$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$ $$\int_{\rm top} \vec Ed\vec A+\int_{\rm bottom} \vec Ed\vec A+\int_{\rm sides} \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$ $$E_1A_{\rm top}+E_3A_{\rm bottom}\cos180^\circ+0=\dfrac{Q_{in}}{\epsilon_0}$$ where $A_{\rm top}=A_{\rm bottom}=A'$ $$E_1A'-E_3A'=\dfrac{Q_{in}}{\epsilon_0}\tag 5$$ Now we need to find the charge enclosed inside our gaussian cylinder where we know that the charge area density is constant for the two slabs. The enclosed charge from the upper slab is given by $$\dfrac{Q}{A}=\dfrac{Q_{\rm in,upper}}{A'}$$ $$Q_{\rm in,upper}=\dfrac{QA'}{A} $$ Thus, $$Q_{in}= \dfrac{QA'}{A} $$ Plug into (5); $$E_1A'-E_3A'=\dfrac{QA'}{A\epsilon_0}$$ $$E_1 -E_3 =\dfrac{Q }{A\epsilon_0}$$ Hence, $$E_3=E_1 -\dfrac{Q }{A\epsilon_0}$$ Plug $E_1$ from the boxed formula above, $$E_3=\dfrac{ 3Q }{2\epsilon_0A}-\dfrac{Q }{A\epsilon_0}$$ $$\boxed{E_3=\dfrac{ Q }{2\epsilon_0A}}$$ $$\color{blue}{\bf [b]}$$ We know that the electric field at the surface of a conductor is given by $$E=\dfrac{\eta}{\epsilon_0}$$ So $$\eta=\epsilon_0E$$ At surface $a$, $$\eta_a=\epsilon_0E_1$$ Plug $E_1$ from the boxed formula above, $$\eta_a=\epsilon_0\dfrac{3Q}{2\epsilon_0 A}$$ $$\boxed{\eta_a= \dfrac{3Q}{2 A}}$$ At surface $b$, $$\eta_b=\epsilon_0E_3$$ Plug $E_3$ from the boxed formula above, and noting that this filed points toward the surface b which means that surface b is negatively charged (polarized). $$\eta_b=\epsilon_0\dfrac{-Q}{2\epsilon_0 A}$$ $$\boxed{\eta_b= \dfrac{-Q}{2 A}}$$ At surface $c$, $$\eta_c=\epsilon_0E_3$$ Plug $E_3$ from the boxed formula above, and noting that this filed points away from the surface c which means that surface c is positively charged (polarized). $$\eta_c=\epsilon_0\dfrac{ Q}{2\epsilon_0 A}$$ $$\boxed{\eta_c= \dfrac{ Q}{2 A}}$$ At surface $d$, $$\eta_d=\epsilon_0E_5$$ Plug $E_5$ from the boxed formula above, and noting that this filed points away from the surface d which means that surface d is positively charged (polarized). $$\eta_d=\epsilon_0\dfrac{ 3Q}{2\epsilon_0 A}$$ $$\boxed{\eta_d= \dfrac{ 3Q}{2 A}}$$