Gauss' law can be relatively easily applied to calculate an electric field of distribution of charges where there's an underlying symmetry. Before you apply Gauss' law to a large sheet of charge, consider the following simplified picture to better understand the relevant symmetry. The uniform charge density +σ allows us to treat (small) equal areas of the sheet as if they were point charges AQ. AQ AQ +o Now consider a point equidistant from the two AQ "point charges" indicated in the figure below. Which of the following is true of the horizontal and vertical components of the electric fields from the "point charges" at the point indicated in black? 0008 The horizontal components will cancel exactly The horizontal components will add and point away from the sheet The vertical components will add and point up The vertical components will add and point down The horizontal components will add and point towards the sheet The vertical components will cancel exactly Write down an expression for the total charge enclosed by the cylindrical Gaussian surface in terms of +σo, r, and Lo Q42 Apply the Gauss' law to find out the magnitude of the electric field E outside the charged sheet. E= σ 200 Which of the following statement(s) are true about the field you just derived? ✓ The electric field outside the large charged sheet does not depend on the distance from the sheet The electric field outside the large charged sheet will be larger if the charge density on it is higher The electric field outside the large charged sheet gets smaller as you move away from the sheet The potential difference between the sheet and any point outside the sheet will depend on the distance to the center of the sheet The electric field outside the large charged sheet will be smaller if the charge density on it is higher The potential difference between the sheet and any point outside the sheet will depend on the distance from the sheet