4Qs AL electrostatic

1. You probably has used the wrong distance in your calculation of potential.
The potential due to a small charge element dq at a distance r from the ring centre is given by,
dV = k.(dq)/square-root(b^2+r^2)
where k is a constant
hence, V = k.Q/square-root(b^2+r^2)
where Q is the total charge on the ring.
substitute r = b and r = 4b respectively into the euqation to calculate V1 and V2, you could find the correct answer.
2. Your method seems right. I think there is some msitake on the given answers.
3. Option A is definitely not the correct answer. I would choose option B.
There are induced +ve and -ve charges on the metal sphere. The external field lines must terminate and originate from these induced charges.
Also, the field lines must be "incident normally" onto the sphere surface as the sphere surface is an equipotential surface.
4. Option B is right. Equipotential lines and field lines are always perpendicluar to each other. If not, there would be a component of the field line along the equipotential line, in this case, the equipotential line will no longer be "equipotential" (because a field line represents a potential gradient, an equipotential line should have zero potential gradient).
Option D is right. Electric field lines are running from high equipotential lines to low equipotential lines. Since a proton carries +ve charges, it will move down along the field lines when free to move, i.e. from places of high potential to places of low potential on its own. Thus if the proton is made to move to a higher potential line, i.e. against the direction of field lines, work has to be done on it to overcome the repulsive force.
Option C is wrong. Electric field strength may be different if the field is not uniform.