The work done to move a charge along an equipotential from \(A\) to \(B\):
1.
can not be defined as \(-\int_{A}^{B} { \vec E\cdot \vec{dl}}\)
2.
must be defined as \(-\int_{A}^{B} {\vec E\cdot \vec{dl}}\)
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:
Step 2: Find the potential difference.
The line integral of the electrical field from the point 1 to 2 gives potential difference.
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.
