Two infinite sheets of charge are separated by \(10.0 \mathrm{~cm}\) as shown in the figure. Sheet 1 has a surface charge distribution of \(\sigma_{1}=3.00 \mu \mathrm{C} / \mathrm{m}^{2}\) and sheet 2 has a surface charge distribution of \(\sigma_{2}=-5.00 \mu \mathrm{C} / \mathrm{m}^{2}\). Find the total electric field (magnitude and direction) at each of the following locations: a) at point \(P, 6.00 \mathrm{~cm}\) to the left of sheet 1 b) at point \(P^{\prime} 6.00 \mathrm{~cm}\) to the right of sheet 1

For an infinite sheet of charge with surface charge density \(\sigma\), the electric field is given by the following formula: $$E = \frac{\sigma}{2 \epsilon_0}$$ Where \(E\) is the electric field in the perpendicular direction of the sheet and \(\epsilon_0\) is the vacuum permittivity constant, with a value of \(8.85 × 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2\).

At point \(P\), which is \(6.00 \mathrm{~cm}\) to the left of sheet 1, both sheets produce electric fields in the same direction (leftwards), as they both have charges with opposite signs. Calculate the electric fields produced by each sheet individually and then combine them by adding the magnitudes of their electric fields at point \(P\). For sheet 1: $$E_1 = \frac{3.00 × 10^{-6} \mathrm{C} / \mathrm{m}^{2}}{2 × 8.85 × 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2} = 169.49 \mathrm{N/C}$$ For sheet 2: $$E_2 = \frac{-5.00 × 10^{-6} \mathrm{C} / \mathrm{m}^{2}}{2 × 8.85 × 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2} = -282.49 \mathrm{N/C}$$ As they are both acting in the same direction (leftwards), the total electric field at point \(P\) will be the sum of their magnitudes: $$E_P = E_1 + E_2 = 169.49 \mathrm{N/C} - 282.49 \mathrm{N/C} = -113.00 \mathrm{N/C}$$ The total electric field at point \(P\) is \(113.00 \mathrm{N/C}\) to the left.

At point \(P'\), which is \(6.00 \mathrm{~cm}\) to the right of sheet 1, both sheets produce electric fields in opposite directions (sheet 1 produces an electric field to the right, and sheet 2 produces an electric field to the left). Calculate the electric fields produced by each sheet individually and then combine them by subtracting the magnitudes of their electric fields at point \(P'\). For sheet 1: $$E_1 = 169.49 \mathrm{N/C}$$ (already calculated in step 2) For sheet 2: $$E_2 = -282.49 \mathrm{N/C}$$ (already calculated in step 2) As they are both acting in opposite directions, the total electric field at point \(P'\) will be the difference between their magnitudes: $$E_{P'} = E_1 - E_2 = 169.49 \mathrm{N/C} - (-282.49 \mathrm{N/C}) = 451.98 \mathrm{N/C}$$ The total electric field at point \(P'\) is \(451.98 \mathrm{N/C}\) to the right.