## 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

1. anonymous

0 V

2. Michele_Laino

an explanation is needed @pinkbubbles

3. anonymous

yes, how did you get that? :/ we need to use a formula?

4. IrishBoy123

" **equi**potential line"

5. anonymous

yes

6. 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

7. Michele_Laino

So*

8. anonymous

ohh and so that is how we get 0 V as the solution?

9. Michele_Laino

yes! we have 0 in voltage change

10. anonymous

Yup!

11. anonymous

thank you!!

12. anonymous

:)

13. Michele_Laino

:)