- #1

- 4,807

- 32

i) V = 0 at z = 0 (since the conducting plane is grounded

ii) V --> 0 far from the charge (that is, for x² + y² + z² >> d²)

My question is: How does grounded implies V = 0?

And how can we have V = 0 at z = 0 and at infinity simultaneously? This would imply, according to [itex]W = Q\Delta V[/itex] that it takes no work to bring a charge Q from infinity to the plane along the z axis. Maybe it IS the case, but it's not obvious at all.

2. This one is for people who own Griffith's book. Look at Figure 2.36 pp.88. How come E_above and E_below point in the same direction ?! I can think of at least one situation where this is not true: for an infinite charged plane, E points outward.