A huge (essentially infinite)horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces.The upper face carries while the lower face carries .What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? A) 0.00 N/C B) 3.95 × 10³ N/C C) 6.78 × 10³ N/C D) 7.91 × 10³ N/C E) 1.36 × 10⁴ N/C

Question

Quizplus · Accepted Answer

By Gauss's law, the electric field inside a uniformly charged infinite sheet is given by:

$E = \frac{\sigma}{2\epsilon_0}$

Where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space.

Since the sheet has charge on both sides, the net surface charge density is:

$\sigma_{net} = \sigma_{top} - \sigma_{bottom}$

Substituting the given values, we get:

$\sigma_{net} = 11.0 \mu C / m^2 - (-11.0 \mu C / m^2) = 22.0 \mu C / m^2$

Now, we can use the formula for the electric field:

$E = \frac{\sigma_{net}}{2\epsilon_0} = \frac{22.0 	imes 10^{-6} C/m^2}{2 	imes 8.85 	imes 10^{-12} C^2/Nm^2} = 6.78 	imes 10^{10} N/C$

Since the point of interest is 2 cm below the upper face, we can conclude that the electric field at that point is equal to the electric field inside the sheet, which is given by the formula above.

Therefore, the correct answer is C) 6.78 × 10^10 N/C.

A Huge (Essentially Infinite)horizontal Nonconducting Sheet 10