In: Physics
Let V = 20x2yz - 10z2 V in free space, (a) Determine the equations of the equipotential surfaces on which V = 0 and 60 V. (b) Assume these are conducting surfaces and find the surface charge density at that point on the V = 60-V surface where x=2 and z= 1. It is known that 0 V 60 V is the field-containing region, (c) Give the unit vector at this point that is normal to the conducting surface and directed toward the V = 0 surface.
(a) The equations of equipotential surfaces is given by the expression:
Then for V=0V and V=60V we can write
And
(b) If we plug the given values of x and z for the V=60V equipotential we get:
On the other hand the E field vector is given by:
If we evaluate it in the point (x,y,z) = (2,7/8,1) we get:
And since the surface charge density is equal to the electric displacement:
(c) The unit normal vector pointing towards lower potential is given by: