Answer :
Answer:
[tex]E_{C} = -5.65 * 10^{11} \hat{x}[/tex] N/C
[tex]E_{A} = -1.24 * 10^{12} \hat{x}[/tex] N/C
Explanation:
The charge per unit area of the two non-conducting slabs are given by:
[tex]\sigma_{a} = -16 C/m^2[/tex]
[tex]\sigma_{b} = 6 C/m^2[/tex]
The charge density on the metal[tex]\sigma_{m} = 0[/tex]
ε0 = 8.854 x 10-12 C2/N m2
Note that the electric field inside the conductor is zero because it is an equipotential surface.
The diagram attached to this solution typifies the description given in the question:
The electric field in the region C can be calculated by:
[tex]E_{C} = \frac{ |\sigma_{b} |- |\sigma_{a}| }{2 \epsilon_{0} } \\E_{C} = \frac{6 - 16 }{2 * 8.854 * 10^{-12} } \\E_{C} = -5.65 * 10^{11} \hat{x}[/tex]
The electric field in the region A can be calculated by:
[tex]E_{A} = \frac{- |\sigma_{a} |- |\sigma_{b}| }{2 \epsilon_{0} } \\E_{A} = \frac{-16 - 6 }{2 * 8.854 * 10^{-12} } \\E_{A} = -1.24 * 10^{12} \hat{x}[/tex]
