Question

A +6.0 mC charge is moved from point A to point B (5.0cm) in a region of space with a uniform electric field of 15 000 N C−1 as shown in the figure above.

What is the change in electrical potential energy of the charge? (to 2 s.f.)

Answer 1

The work that is needed to move the charge can be calculated as

$A_{12} = \vec{E}\cdot q\cdot \vec{\Delta l} = 15000\,\dfrac{\mathrm{N}}{\mathrm{C}}\cdot 6.0\cdot10^{-3}\,\mathrm{C}\cdot0.05\,\mathrm{m}\cdot\cos\alpha =4.5\,\mathrm{J}\cdot\cos\alpha ,$

where $\alpha$ is the angle between the direction of electric force and the displacement. Work is equal to the difference of potential energies:

$A_{12} = E_{1} - E_{2},$

so we can say that the change of energy is strongly dependent on the angle of displacement and

$E_2 -E_1 = -4.5\,\mathrm{J}\cdot\cos\alpha,$

where $\alpha$ should be taken from the full formulation of the problem.