Question Solved1 Answer need help with number 3 parts d and e please 3. Two very large sheets of charge are separated by a distance d. One sheet has a surface charge density +o, and the other a surface charge density -0.. A small region near the center of the sheets is shown. a. Draw arrows on the diagram to indicate the direction of the electric field at points A, B, C, and D. i. Compare the magnitudes of the electric fields at points A, B, C, and D. Explain. Electric field at the points are the same because the value of E' does not depend on distance B C D E d. Find the magnitude and direction of the electric field at points A, B, C, and D. (Hint: Use superposition and your results for the electric field of a large sheet from the Gauss' law tutorial.) e. A particle of mass m, and charge -9. is released from rest at a point just to the left of the right sheet. Find the speed of the particle as it reaches the left sheet.

Here, as we are going to have the magnitude and direction of the electric field at points A,B,C and D in part (d), so for part (a), only qualitative discussion is asked. and part (d) will cover it mathematically as per requirement of the question.   So,   for part (a), we have the directions of electric field at points A, B, C and D as given below. UNIFORM ELECTRIC FIELD Fig 1. Representing uniform electric field between charged sheets Fig 2. Representing the direction of electric fields at Point A, B, C, and D     in the above figure, the first figure represents the general direction of electric field between those plates and the second figure gives the direction at points A,B,C and D. The electric field direction comes out as given in figure because as the electic field always emerges from positive charge and sinks at negative charge. Positive charge is the source of electric field and negative charge is the sink for the same.     So, in this set of parallel plates, the direction of electric field is normal to the charged surface and and goes straight from the sheet having positive charge to the sheet having negative charge.       Part (i): Here, As the property of electric field lines suggest the fact that the electric field lines always try to go radially outwards unless there is some non radial convergence due to some sink nearby, so as the fig.1 suggests the uniform electric field throughout due to symmetrical distribution of charges providing the emerging field lines a path to go straight into sink rather than converging like in fig.3 below, the electric field lines are uniform throughout. Therefore, the electric field intensity remains constant all over the region between those sheets as the electric field intensity is directly proportional to the number of electric field lines passing thorugh the region. This will be prooven mathematically in part (d) as per the requirement of the question.     Fig 3. showing the curving of electric field in presence of opposite charge near it's vicinity       Part (d): Here, Another figure is drawn to give a simple representation of the Scenario. here, the are 2 uniformly charged plates having surface charge densities as +sigma_(0) and -sigma_(0) for positive and negative charge respectively.       Then, according to gauss's law, we have: ointE.ds=(Q)/(epsilon_(0))     Where: Q= Total Charge   E= Electric Field   epsilon_(0)= Electrical permitivity of free space =8.854 xx10^(-12)F//m     And the integeral is over the surface. this gives us: E.2ointds=(sigma_(0)ointds)/(epsilon_(0))       Here, Q=sigma_(0)ointds because sigma is represented ... See the full answer