Question

Two infinite, uniformly charged, flat, nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2},\) and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

Short answer

Answer: The magnitude of the net electric field at any point not on either of the charged surfaces is approximately \(3584.9 \ \mathrm{N} / \mathrm{C}\).

Step-by-step solution

1. Electric field due to an infinite uniformly charged flat surface

According to Gauss's law, the electric field (E) due to an infinite uniformly charged flat surface with surface charge density σ is given by: \(E = \frac{\sigma}{2 \epsilon_0}\), where ε₀ is the vacuum permittivity.

2. Calculate the electric field due to surface with positive charge distribution

Given the positive surface charge distribution is \(+30.0 \ \mathrm{pC} \ / \mathrm{m}^2\). We convert this to Coulombs per square meter: \(+(30.0 * 10^{-12}) \ \mathrm{C} \ / \mathrm{m}^2\). The electric field due to this surface will be: \(E_1 = \frac{(30.0 * 10^{-12})}{2 \epsilon_0} \)

3. Calculate the electric field due to surface with negative charge distribution

Given the negative surface charge distribution is \(-40.0 \ \mathrm{pC} \ / \mathrm{m}^2\). We convert this to Coulombs per square meter: \(-(40.0 * 10^{-12}) \ \mathrm{C} \ / \mathrm{m}^2\). The electric field due to this surface will be: \(E_2 = \frac{(-40.0 * 10^{-12})}{2 \epsilon_0} \)

4. Apply superposition principle to find the total electric field

The principle of superposition states that the net electric field at any point is equal to the vector sum of the electric fields due to individual charged surfaces. Since the surfaces are mutually perpendicular, the electric fields are perpendicular to each other too. Thus, the magnitude of the net electric field (E) is given by: \(E = \sqrt{E_{1}^{2} + E_{2}^{2}}\)

5. Calculate the magnitude of the net electric field

Now plug in the values for \(E_1\) and \(E_2\) we obtained from steps 2 and 3: \(E = \sqrt{ \left( \frac{(30.0 * 10^{-12})}{2 \epsilon_0} \right)^{2} + \left( \frac{(-40.0 * 10^{-12})}{2 \epsilon_0} \right)^{2} }\) After substitution of the vacuum permittivity value (\(\epsilon_0 = 8.85 * 10^{-12} \ \mathrm{F} / \mathrm{m}\)), we get: \(E = \sqrt{ \left( \frac{(30.0 * 10^{-12})}{(2 * 8.85 * 10^{-12})} \right)^{2} + \left( \frac{(-40.0 * 10^{-12})}{(2 * 8.85 * 10^{-12})} \right)^{2} }\) Finally, after evaluating the expression, we find the magnitude of the net electric field: \(E \approx 3584.9 \ \mathrm{N} / \mathrm{C}\)