Answer to Question #116624 in Electric Circuits for Jess

Question #116624
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.)
1
Expert's answer
2020-05-18T10:24:42-0400

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

A12=EqΔl=15000NC6.0103C0.05mcosα=4.5Jcosα,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 α\alpha is the angle between the direction of electric force and the displacement. Work is equal to the difference of potential energies:

A12=E1E2,A_{12} = E_{1} - E_{2},

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

E2E1=4.5Jcosα,E_2 -E_1 = -4.5\,\mathrm{J}\cdot\cos\alpha,

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


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