Electric field in insulating slab

• temaire
In summary, the conversation discusses the process of determining the electric field at a specific point within a large, thin, insulating slab with a uniform charge distribution. Using Gauss' Law and the concept of Gaussian surfaces, the net electric field at the point is calculated to be 2.26 N/C. The process involves dividing the slab into two pillbox-shaped Gaussian surfaces and calculating the electric field from each surface. The final answer is confirmed to be correct, except for a minor correction in terminology.
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?

temaire said:
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.

1 person

1. What is an insulating slab?

An insulating slab is a material that does not conduct electricity easily. This means that it has a high resistance to the flow of electric current and does not allow the movement of electrons through it.

2. What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. It is created by the presence of electric charges and can either attract or repel other charges.

3. How is an electric field created in an insulating slab?

An electric field in an insulating slab is created by the presence of electric charges on its surface. These charges create an electric potential difference between the two sides of the slab, resulting in an electric field within the slab.

4. What is the direction of the electric field in an insulating slab?

The direction of the electric field in an insulating slab is always perpendicular to its surface. This means that the electric field lines are parallel to the surface of the slab and do not intersect it.

5. How does the thickness of an insulating slab affect the electric field?

The thickness of an insulating slab has a direct effect on the strength of the electric field within it. The thicker the slab, the weaker the electric field will be. This is because the electric charges on the surface of the slab are spread out over a larger area, resulting in a lower electric potential difference and weaker electric field.

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