Question

The electric field in a region is given by \( \vec{E} = 40x \hat{i} \, \text{N/C}. \) Find the amount of work done in taking a unit positive charge from a point (0, 3m) to the point (5m, 0).

Answer 1

We are given the electric field \( \vec{E} = 40x \hat{i} \, \text{N/C} \), where \( x \) is the position in the x-direction. The task is to calculate the work done in moving a unit positive charge from the point (0, 3m) to the point (5m, 0).

The work done \( W \) in moving a charge \( q \) in an electric field \( \vec{E} \) is given by the line integral:

\[ W = \int \vec{F} \cdot d\vec{r} \]

Where \( \vec{F} = q\vec{E} \) is the force acting on the charge. For a unit positive charge, \( q = 1 \). Hence, the work done is:

\[ W = \int_{(0, 3)}^{(5, 0)} \vec{E} \cdot d\vec{r} \]

Since the electric field \( \vec{E} \) is along the x-axis and only depends on \( x \), we can write the displacement vector \( d\vec{r} \) as:

\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]

Substitute the components of \( \vec{E} = 40x \hat{i} \) into the equation for work:

\[ W = \int_{0}^{5} (40x) \, dx \]

Now, integrating:

\[ W = \left[ 20x^2 \right]_0^5 = 20(5^2) - 20(0^2) = 20(25) = 500 \, \text{J} \]

Therefore, the work done in moving the unit positive charge from the point (0, 3m) to the point (5m, 0) is \( 500 \, \text{J} \).