33Yes i forgot to mention... V is constant throughout the right side.
But see the answer again.
The left side of the plane has 0 potential and right has a positive potential.
a positive charge is incident at boundary as shown with sufficient velocity as to enter right region. The excepted path in right side is
I
II
III
IV
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7 Answers
11II(assuming constant potential)
check the answer as well as assumption :P
3311constant potential so zero field
then why does it turn my ans is still II
33At first look it seems II but think again.
...with sufficient velocity as to enter right region....
106dV/dr gives component of electric field along the path and as it is zero so it experiences no external force... due to field... so II??
33This question??
E=-dV/dx
You all have forgot that V is uniform but changes as we pass the line so dV=V as we pass the line.
now V is positive so <----- (direction of E at boundary)
and q is +ve so III