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

To calculate the work done by the electric field in moving a charge, we use the line integral of the force over the path taken. The work done $W$ is given by:
$$W = \int \vec{F} \cdot d\vec{r}$$
Since the force on a charge $q$ in an electric field $\vec{E}$ is $\vec{F} = q\vec{E}$, the work done is:
$$W = q \int \vec{E} \cdot d\vec{r}$$
Given: $q = 3 \text{ C}$, $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}$, and $d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}$.

The integral becomes:
$$W = 3 \int_{(0, -2, -5)}^{(5, 1, 2)} (2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k})$$
$$W = 3 \left[ \int_{0}^{5} 2x dx + \int_{-2}^{1} 3y^2 dy + \int_{-5}^{2} 4 dz \right]$$

Integrating each term separately:
1. $\int_{0}^{5} 2x dx = [x^2]_0^5 = 25 - 0 = 25$
2. $\int_{-2}^{1} 3y^2 dy = [y^3]_{-2}^1 = 1^3 - (-2)^3 = 1 - (-8) = 9$
3. $\int_{-5}^{2} 4 dz = [4z]_{-5}^2 = 4(2) - 4(-5) = 8 + 20 = 28$

Summing these results:
$$W = 3 [ 25 + 9 + 28 ]$$
$$W = 3 [ 62 ] = 186 \text{ J}$$
Therefore, the work done in moving the charge is $186 \text{ J}$.