Question
A very large nonconducting plate lying in the xy-plane carries a charge per unit area of 8s. A second such plate located at z = 5.00 cm and oriented parallel to the xy-plane carries a charge per unit area of -9s. Find the electric field for the following.
(a) z < 0
___________s/e0
(b) 0 < z < 5.00 cm
___________s/e0
(c) z > 5.00 cm
___________s/e0
Explanation / Answer
, for infite sheet of charge with charge density s, the electric field will be given by E = s / (8 * eo). As you can see, the value is not dependent on the distance of the point from the sheet. The direction will depend on the nature of charge of the sheet and the location of the point. In this case, one sheet has a charge density of 8s (implies it has +ve charge). Hence the electric field will be pointed away from the sheet. Also this is placed at z = 0, on xy plane. Hence the electric field direction due to this sheet will be away from the sheet, i.e., towards -ve z axis for points on -ve z-axis and towards +ve z axis for points on +ve z-axis. For second sheet charge density is -9s (implies it has -ve charge). Hence the electric field will be pointed into the sheet. Also this is placed at z = 5, on xy plane. Hence the electric field direction due to this sheet will be towards +ve z axis for points on on z-axis with z < 5 and towards +ve z axis for points on +ve z-axis for z>5. The resultant field is just the sum of [8s / (8 * eo)] and [9s / (8 * eo)] considering the directions of the fields at the required points. Say for points z < 0, it will be -[8s / (8 * eo)] + [9s / (8 * eo)] = 1s / (8* eo). for points 0 5 it will be [8s / (8 * eo)] - [9s / (8* eo)] = -1s / (8 * eo). In the above values, +ve value implies field along +ve z-axis and -ve sign implies field along -ve z-axis. Calculate the values of E for the points you needed . Hope this helps you.