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anonymous
 one year ago
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
anonymous
 one year ago
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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1an explanation is needed @pinkbubbles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, how did you get that? :/ we need to use a formula?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0" **equi**potential line"

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1equipotential 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh and so that is how we get 0 V as the solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we have 0 in voltage change
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