A three coulomb charge moves from the point (0, -2, -5) to the...

The force experienced by the charge in the electric field is given by $\vec{F} = q\vec{E}$.

Given $q = 3$ C and $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}$ N/C, the force is:
$\vec{F} = 3(2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) = 6x\hat{i} + 9y^2\hat{j} + 12\hat{k}$ N

The work done by the electric field in moving the charge is:
$W = \int \vec{F} \cdot d\vec{r} = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz$

Substituting the limits from the initial point $(0, -2, -5)$ to the final point $(5, 1, 2)$:
$W = \int_{0}^{5} 6x dx + \int_{-2}^{1} 9y^2 dy + \int_{-5}^{2} 12 dz$

Evaluating each integral separately:
$\int_{0}^{5} 6x dx = \left[ 3x^2 \right]_{0}^{5} = 3(5^2 - 0^2) = 75$ J

$\int_{-2}^{1} 9y^2 dy = \left[ 3y^3 \right]_{-2}^{1} = 3(1^3 - (-2)^3) = 3(1 - (-8)) = 3(9) = 27$ J

$\int_{-5}^{2} 12 dz = \left[ 12z \right]_{-5}^{2} = 12(2 - (-5)) = 12(7) = 84$ J

Total work done $W = 75 + 27 + 84 = 186$ J.

Answer: $186$