11constant potential so zero field
then why does it turn my ans is still II
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
33Yes i forgot to mention... V is constant throughout the right side.
But see the answer again.
1133At 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