Work Done Moving Charge in Non Uniform Electric Field

To find the work done in moving a charge in an electric field, we use the definition of work done by an electric force:

$W = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} = q \int_{\vec{r}_i}^{\vec{r}_f} \vec{E} \cdot d\vec{r}$

Given data:

• Charge, $q = 3\text{ C}$
• Electric field, $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}\text{ N/C}$
• Initial point, $\vec{r}_i = (0, -2, -5)$
• Final point, $\vec{r}_f = (5, 1, 2)$

The differential displacement vector $d\vec{r}$ is given by: $d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}$

Taking the dot product $\vec{E} \cdot d\vec{r}$: $\vec{E} \cdot d\vec{r} = 2x\,dx + 3y^2\,dy + 4\,dz$

Now, integrating from the initial point to the final point: $\int_{\vec{r}_i}^{\vec{r}_f} \vec{E} \cdot d\vec{r} = \int_{0}^{5} 2x\,dx + \int_{-2}^{1} 3y^2\,dy + \int_{-5}^{2} 4\,dz$

Evaluating the integrals individually:

1. $x$-component: $\int_{0}^{5} 2x\,dx = \left[ x^2 \right]_{0}^{5} = 5^2 - 0^2 = 25$

2. $y$-component: $\int_{-2}^{1} 3y^2\,dy = \left[ y^3 \right]_{-2}^{1} = 1^3 - (-2)^3 = 1 - (-8) = 9$

3. $z$-component: $\int_{-5}^{2} 4\,dz = \left[ 4z \right]_{-5}^{2} = 4(2) - 4(-5) = 8 + 20 = 28$

Summing the results of the integrals: $\int_{\vec{r}_i}^{\vec{r}_f} \vec{E} \cdot d\vec{r} = 25 + 9 + 28 = 62$

Now, calculating the work done $W$: $W = q \left( \int_{\vec{r}_i}^{\vec{r}_f} \vec{E} \cdot d\vec{r} \right) = 3 \times 62 = 186\text{ J}$

Thus, the work done in moving the charge is $186\text{ J}$.