Electric field in insulating slab

[temaire]

A large, thin, insulating slab 2 m x 2 m x 5 mm has a charge of 2 x 10^{-10} C distributed uniformly throughout its volume. determine the electric field at observation point P, which is located within the slab, beneath its centre, 0.5 mm from the top face.

I started off by making two Gaussian surfaces in the shape of cylinders. The first ranges from the bottom of the slab to point P. The second ranges from point P to the top of the slab. I then calculated the net electric field on point P due to the electric fields traveling out of the ends of the two Gaussian surfaces, using Gauss' Law.

E = E_{1} - E_{2} = \frac{Q_{1}}{2A\epsilon_{o}} - \frac{Q_{2}}{2A\epsilon_{o}}

where E is the net electric field acting on point P, E_{1} is the electric field from the first Gaussian surface, and E_{2} is the electric field from the second Gaussian surface.

The charge density is as follows:

\rho = Q/V = 2\cdot 10^{-10}/(2\cdot2\cdot0.005) = 1\cdot10^{-8}

Q_{1}=\rho V = \rho A \cdot 0.0045 = 4.5\cdot 10^{-11}A

Q_{2}=\rho V = \rho A \cdot 0.0005 = 5\cdot 10^{-12}A

E = \frac{4.5\cdot 10^{-11}A}{2A\epsilon_{o}} - \frac{5\cdot 10^{-12}A}{2A\epsilon_{o}}

E = 2.26 N/C

My final answer is 2.26 N/C. Is my process correct?

 

[collinsmark]

A large, thin, insulating slab 2 m x 2 m x 5 mm has a charge of 2 x 10^{-10} C distributed uniformly throughout its volume. determine the electric field at observation point P, which is located within the slab, beneath its centre, 0.5 mm from the top face.

I started off by making two Gaussian surfaces in the shape of cylinders. The first ranges from the bottom of the slab to point P. The second ranges from point P to the top of the slab. I then calculated the net electric field on point P due to the electric fields traveling out of the ends of the two Gaussian surfaces, using Gauss' Law.

E = E_{1} - E_{2} = \frac{Q_{1}}{2A\epsilon_{o}} - \frac{Q_{2}}{2A\epsilon_{o}}

where E is the net electric field acting on point P, E_{1} is the electric field from the first Gaussian surface, and E_{2} is the electric field from the second Gaussian surface.

The charge density is as follows:

\rho = Q/V = 2\cdot 10^{-10}/(2\cdot2\cdot0.005) = 1\cdot10^{-8}

Q_{1}=\rho V = \rho A \cdot 0.0045 = 4.5\cdot 10^{-11}A

Q_{2}=\rho V = \rho A \cdot 0.0005 = 5\cdot 10^{-12}A

E = \frac{4.5\cdot 10^{-11}A}{2A\epsilon_{o}} - \frac{5\cdot 10^{-12}A}{2A\epsilon_{o}}

E = 2.26 N/C

My final answer is 2.26 N/C. Is my process correct?

That looks correct to me.

Oh, except for one thing. You said, "Gaussian surfaces in the shape of cylinders." I think you meant pillboxes.