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…

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

Given $q = 3$q=3 C and $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}$E=2xi^+3y2j^​+4k^ N/C, the force is:

$\vec{F} = 3(2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) = 6x\hat{i} + 9y^2\hat{j} + 12\hat{k}$F=3(2xi^+3y2j^​+4k^)=6xi^+9y2j^​+12k^ N

The work done by the electric field in moving the charge is:

$$W = \int \vec{F} \cdot d\vec{r} = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz$$W=∫F⋅dr=∫x1​x2​​Fx​dx+∫y1​y2​​Fy​dy+∫z1​z2​​Fz​dz

Substituting the limits from the initial point $(0, -2, -5)$(0,−2,−5) to the final point $(5, 1, 2)$(5,1,2):

$$W = \int_{0}^{5} 6x dx + \int_{-2}^{1} 9y^2 dy + \int_{-5}^{2} 12 dz$$W=∫05​6xdx+∫−21​9y2dy+∫−52​12dz

Evaluating each integral separately:

$\displaystyle \int_{0}^{5} 6x dx = \left[ 3x^2 \right]_{0}^{5} = 3(5^2 - 0^2) = 75$∫05​6xdx=[3x2]05​=3(52−02)=75 J

$\displaystyle \int_{-2}^{1} 9y^2 dy = \left[ 3y^3 \right]_{-2}^{1} = 3(1^3 - (-2)^3) = 3(1 - (-8)) = 3(9) = 27$∫−21​9y2dy=[3y3]−21​=3(13−(−2)3)=3(1−(−8))=3(9)=27 J

$\displaystyle \int_{-5}^{2} 12 dz = \left[ 12z \right]_{-5}^{2} = 12(2 - (-5)) = 12(7) = 84$∫−52​12dz=[12z]−52​=12(2−(−5))=12(7)=84 J

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

Answer: $186$186