Question

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

Hint:

Work done is calculated as the integral of $q\vec{E} \cdot d\vec{r}$. Solve the integral for x, y, and z separately using the given start and end coordinates.

Answer 1

Correct Answer: 186

Work done by an external agent in an electric field is the negative change in potential energy, but here we calculate the work done by the field itself or simply evaluate the integral $W = \int \vec{F} \cdot d\vec{l}$.

The force is $\vec{F} = q\vec{E} = 3(2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) = 6x\hat{i} + 9y^2\hat{j} + 12\hat{k}$.
The displacement vector is $d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}$.
The work done is $W = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz$.

Step-by-step integration:
Along the x-axis: $\int_0^5 6x dx = [3x^2]_0^5 = 3(25) - 0 = 75 \text{ J}$.
Along the y-axis: $\int_{-2}^1 9y^2 dy = [3y^3]_{-2}^1 = 3(1^3 - (-2)^3) = 3(1 + 8) = 27 \text{ J}$.
Along the z-axis: $\int_{-5}^2 12 dz = [12z]_{-5}^2 = 12(2 - (-5)) = 12(7) = 84 \text{ J}$.

Total work done $W = 75 + 27 + 84 = 186 \text{ J}$.