anonymous
  • anonymous
A 1.0 µC test charge travels along an equipotential line a distance of 0.20 cm between two parallel charged plates with a field strength of 500.0 N/C. What is the change in voltage? (µC = 1.0 × 10^-6 C) A. -100 V B. 0 V C. +100 V D. +1,000 V
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
0 V
Michele_Laino
  • Michele_Laino
an explanation is needed @pinkbubbles
anonymous
  • anonymous
yes, how did you get that? :/ we need to use a formula?

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IrishBoy123
  • IrishBoy123
" **equi**potential line"
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
equipotential line is a line for which the electric potential is constant. SO during its motion our charge doesn't change its potential. Now we have this formula: \[\Delta V = \frac{{\Delta U}}{q}\] so if \Delat U is zero, then also \Delta V is zero
Michele_Laino
  • Michele_Laino
So*
anonymous
  • anonymous
ohh and so that is how we get 0 V as the solution?
Michele_Laino
  • Michele_Laino
yes! we have 0 in voltage change
anonymous
  • anonymous
Yup!
anonymous
  • anonymous
thank you!!
anonymous
  • anonymous
:)
Michele_Laino
  • Michele_Laino
:)

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