A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as E = 2x i + 3y^2 j + 4 k N/C. The work done in moving the charge is _______ J.

Step-by-step solution

The force experienced by the charge in the electric field is given by F = q E . Given q = 3 C and E = 2x i + 3y^2 j + 4 k N/C, the force is: F = 3(2x i + 3y^2 j + 4 k ) = 6x i + 9y^2 j + 12 k N The work done by the electric field in moving the charge is: W = F d r = _ x₁ ^ x₂ F_x dx + _ y₁ ^ y₂ F_y dy + _ z₁ ^ z₂ F_z dz Substituting the limits from the initial point (0, -2, -5) to the final point (5, 1, 2) : W = ₀⁵ 6x dx + _ -2 ¹ 9y^2 dy + _ -5 ² 12 dz Evaluating each integral separately: ₀⁵ 6x dx = [ 3x^2 ]₀⁵ = 3(