Question

A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb. The distance between A and B is 1.2 × 10^-2 meters and the path between A and B is parallel to the field. What is the magnitude of the difference in potential energy?
A. 1.2 × 10^-15 joules
B. 2.3 × 10^-15 joules
C. 3.2 × 10^-15 joules
D. 6.4 × 10^-15 joules
E. 6.4 × 10^-14 joules

Answer 1

In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb

We have the magnitude of the load:

q = 6.4 × $10 ^{-19}$ coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × $10^{-2}$ meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × $10^{-19}$)(6.5 × $10^4$)(1.2 × $10^{-2}$)

PE = 5.0 x $10^{-16}$ joules

None of the options shown is correct.