Question

A particle moves through a distance of 10 m along an electric field of strength 75 N/C. Its electrical potential energy decreases by 4.8 x 10-16 J. What is the potential difference between the initial and final locations of the particle?


750 V

-750 V

75 V

-75 V

Answer 1

Answer is -750 V

Answer 2

here we have to apply the conservation of total energy, namely since the electric field is a conservative field, then the sum of kinetic energy and potential energy has to be constant:
\[\Large \frac{1}{2}mv_1^2 + {U_1} = \frac{1}{2}mv_2^2 + {U_2}\]

Answer 3

where m is the mass of our particle.
Now the work W done by the electric field, is given by the subsequent formula:
\[\Large W = qEd = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = {U_1} - {U_2}\]
wher q is the electric charge of our particle

Answer 4

so we can write this:
\[\Large \Delta V = Ed = \frac{{{U_1} - {U_2}}}{q}\]

Answer 5

@Michele_Laino thank you so much!