Question

The electric field in a region is given by: \[ \vec{E} = (10x + 4) \hat{i} \] where \( x \) is in meters and \( \vec{E} \) is in N/C. Calculate the amount of work done in taking a unit charge from: (i) \( (5 m, 0) \) to \( (10 m, 0) \) (ii) \( (5 m, 0) \) to \( (5 m, 10 m) \)

Hint:

Work done in an electric field depends on displacement along the field direction. No work is done if the movement is perpendicular to the field.

Answer 1

Work Done in Moving a Unit Charge 

The work done in moving a charge \( q \) in an electric field is given by: \[ W = \int_{x_1}^{x_2} q E \, dx \] For a unit charge (\( q = 1 \)), this simplifies to: \[ W = \int_{x_1}^{x_2} E \, dx \] (i) Work Done from \( (5 m, 0) \) to \( (10 m, 0) \)
Since the electric field is along the \( x \)-axis, we compute: \[ W = \int_{5}^{10} (10x + 4) \, dx \] \[ W = \left[ 10 \frac{x^2}{2} + 4x \right]_{5}^{10} \] \[ W = \left( 5 \times 100 + 4 \times 10 \right) - \left( 5 \times 25 + 4 \times 5 \right) \] \[ W = (500 + 40) - (125 + 20) \] \[ W = 540 - 145 = 395 \text{ J} \] Thus, the work done is 395 J. 
(ii) Work Done from \( (5 m, 0) \) to \( (5 m, 10 m) \) - Since the electric field is only along the \( x \)-direction (\( E_x \)), there is no electric field component in the \( y \)-direction.
- Work is only done when moving in the direction of the field. Since displacement in the \( x \)-direction is zero, the work done is: \[ W = 0 \] Thus, the work done is 0 J.