An electric field is given as E = 6y2z i + 12xyz j + 6xy2 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)

To calculate the work done in moving a charge in an electric field, we use the formula:

[ W = q int vec{E} cdot dvec{l} ]

Where:

– ( W ) is the work done,

– ( q ) is the charge (in Coulombs),

– ( vec{E} ) is the electric field,

– ( dvec{l} ) is the differential path element.

First, we need to evaluate ( vec{E} ) at the point ( p(0, 2, 5) ):

[

E = 6y^2z hat{i} + 12xyz hat{j} + 6xy^2 hat{k}

]

Substituting ( y = 2 ) and ( z = 5 ):

1. For the ( hat{i} ) component:

[

E_x = 6(2^2)(5) = 6(4)(5) = 120

]

2. For the ( hat{j} ) component:

[

E_y = 12(0)(2)(5) = 0

]

3. For the ( hat{k} ) component:

[

E_z = 6(0)(2^2) = 0

]

Thus, at point