Question

An electric field is given as E = 6y^2z i + 12xyz j + 6xy^2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule)

(a) 0.64

(b) 0.72

(c) 0.78

(d) 0.80

The question was asked by my school teacher while I was bunking the class.

This question is from Line Integral topic in section Vector Calculus of Electromagnetic Theory

Answer 1

The correct option is (b) 0.72

Easy explanation: W = -Q E.dl

W = -2 X 10^-3 X (6y^2z i + 12xyz j + 6xy^2 k) . (-3 i + 5 j -2 k)

At p(0,2,5), W = -2(-18.22.5) X 10^-3 = 0.72 J.