anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Answer is -750 V
Michele_Laino
  • 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: \[\Large \frac{1}{2}mv_1^2 + {U_1} = \frac{1}{2}mv_2^2 + {U_2}\]
Michele_Laino
  • 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

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Michele_Laino
  • Michele_Laino
so we can write this: \[\Large \Delta V = Ed = \frac{{{U_1} - {U_2}}}{q}\]
anonymous
  • anonymous
@Michele_Laino thank you so much!

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