Solution:
Work done in a conservative field is not dependent on path taken by the object so,
\[ dV= - \overrightarrow{E}.\overrightarrow{dr}\]
\[ V= - E\widehat{i}.(-a\widehat{i}-b\widehat{j})\]
so, Potential difference = -E.a
Work Done= -Q. E.a
Point charge (q) moves from point (P) to point (S) along the path PQRS as shown in fig. in a uniform electric field E pointing co parallel to the positive direction of the x-axis. The co-ordinates of the points P,Q,R and S are (a, b, 0), (2a, 0, 0), (a, –b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression
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