Question

A charge of 1.5 × 10^15 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10^-5 newtons/coulomb and the distance along the field lines between the two points is 1.7 × 10^-2 meters. What is the work done? How do you firgure this out? please help !

Answer 1

Does anyone know? No teachers at my school teach physics..... and its my last class i have to finish to graduate

Answer 2

For a point charge moving in a uniform electric field (parallel plate capacitor), the work done by the field on the charge is equal to the product of the force applied by the field, and the distance traveled by the charged body. I.e. W=q*E*d, where W=work, q=charge, E=electric field. Looks like your answer will simply be the product of the three real numbers you gave.

Answer 3

*And d=distance traveled by the charge.

Answer 4

W = F d = q E d = 1.5E-15 * 2.5E-5 * 1.7E-2 = 6.4 * 10^-22 J

Answer 5

W= F*d = E*q*d

Answer 6

thank you so much that helped so much!