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