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b
Explanation: W = -Q E.dl
W = -2 X 10-3 X (6y2z i + 12xyz j + 6xy2 k) . (-3 i + 5 j -2 k)
At p(0,2,5), W = -2(-18.22.5) X 10-3 = 0.72 J.
To find the work done in moving a charge (q) along an incremental path (mathbf{dl}) in an electric field (mathbf{E}), the formula used is:
[ dW = q mathbf{E} cdot mathbf{dl} ]
Given:
– The electric field (mathbf{E} = 6y^2z mathbf{i} + 12xyz mathbf{j} + 6xy^2 mathbf{k}).
– The incremental path (mathbf{dl} = -3 mathbf{i} + 5 mathbf{j} – 2 mathbf{k}) mm, or converting to meters (since the electric field is in standard SI units) gives us (mathbf{dl} = -0.003 mathbf{i} + 0.005 mathbf{j} – 0.002 mathbf{k}) meters.
– The charge (q = 2 mC = 2 times 10^{-3} C).
– The location of the path is at (p(0,2,5)).
First, evaluate the electric field (mathbf{E}) at point (p(0,2,5)):
[ mathbf{E}(0,2,5) = 6(2)^2(5) mathbf{i} +