The work done to move a charge along an equipotential from A to B:

1. can not be defined as \(-\int_{\mathrm{A}}^{\mathrm{B}} \text { E. dl. }\)
2. must be defined as \(-\int_{\mathrm{A}}^{\mathrm{B}} \text { E. dl. }\)
3. is zero
4. can have a non-zero value.

NEETprep Answer: 
(3) Hint: The work done depends on the potential difference.
Step 1: Find the work done.
Work done in displacing a charged particle is given by:
W12=q(V2-V1)
Step 2: Find the potential difference.
The line integral of the electrical field from the point 1 to 2 gives potential difference.
V2-V1=-12E.dl
For equipotential surface, V2-V1 = 0 and W = 0
Note: If the displaced charged particle is + 1 C, then and only then, option (b) is correct. But the NCERT exemplar book has given (b) as the correct option which is probably not so under the given conditions.