The electric field near the surfaceof a uniformly charged conducting plane can b

Question

The electric field near the surfaceof a uniformly charged conducting plane can be determined byGauss's law. If the plane is a square of sides 19.0 cm and carriesa charge of 32.5 nC, what is the magnitude of the electric fieldnear the surface?.... **(cont.) Now let's say that aplane with a charge of equal magnitude but opposite sign is broughtnear the first. What is the force between the twoplanes? The electric field near the surfaceof a uniformly charged conducting plane can be determined byGauss's law. If the plane is a square of sides 19.0 cm and carriesa charge of 32.5 nC, what is the magnitude of the electric fieldnear the surface?.... **(cont.) Now let's say that aplane with a charge of equal magnitude but opposite sign is broughtnear the first. What is the force between the twoplanes?

Explanation / Answer

           Given               charge on the square   q = 32.5 nC = 32.5*10-9 C                  area of the square A = (19*19 )cm2                                                    = 0.361 m2                             the surface charge density                                     = Q / A                                       = Q / L2                                     = (32.5*10-9) / (0.19)2                                        =900*10-9 C/m2 the magnitude of the electric field near the surface                              E   =   /20                   = (900*10-9) / 2(8.85*10-12)                   = 50.8*103 N/C                             the surface charge density                                     = Q / A                                       = Q / L2                                     = (32.5*10-9) / (0.19)2                                        =900*10-9 C/m2 the magnitude of the electric field near the surface                              E   =   /20                   = (900*10-9) / 2(8.85*10-12)                   = 50.8*103 N/C                                     = Q / A                                       = Q / L2                                     = (32.5*10-9) / (0.19)2                                        =900*10-9 C/m2 the magnitude of the electric field near the surface                              E   =   /20                   = (900*10-9) / 2(8.85*10-12)                   = 50.8*103 N/C

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