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

Given an electric field \( \vec{E} = 40x \hat{i} \, \text{N/C} \), where \( x \) denotes the position along the x-axis, we aim to compute the work done when moving a unit positive charge from (0, 3m) to (5m, 0).

The work \( W \) required to move a charge \( q \) in an electric field \( \vec{E} \) is defined by the line integral:

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

The force \( \vec{F} \) on the charge is \( q\vec{E} \). For a unit positive charge, \( q = 1 \), so the work done simplifies to:

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

As the electric field \( \vec{E} \) is unidirectional along the x-axis and is a function of \( x \) only, the differential displacement vector \( d\vec{r} \) can be expressed as:

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

Substituting the components of \( \vec{E} = 40x \hat{i} \) into the work integral equation yields:

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

Performing the integration:

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

Consequently, the work expended to transport the unit positive charge from (0, 3m) to (5m, 0) is \( 500 \, \text{J} \).