## anonymous one year ago 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

1. anonymous

2. Michele_Laino

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: $\Large \frac{1}{2}mv_1^2 + {U_1} = \frac{1}{2}mv_2^2 + {U_2}$

3. Michele_Laino

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

4. Michele_Laino

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

5. anonymous

@Michele_Laino thank you so much!