Ncert Exercises & Solutions

The work done to move a charge along an equipotential from

A

$A$ to

B

$B$ :

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:

W_{12} = q (V_{2} - V_{1})

$W_{12} = q (V_{2} - V_{1})$

Step 2: Find the potential difference.

The line integral of the electrical field from the point 1 to 2 gives potential difference.

V_{2} - V_{1} = - \int_{1}^{2} E . dl

$V_{2} - V_{1} = - \int_{1}^{2} E . dl$

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.

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