Question

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

Answer 1

0 V

Answer 2

an explanation is needed @pinkbubbles

Answer 3

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

Answer 4

" **equi**potential line"

Answer 5

yes

Answer 6

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

Answer 7

So*

Answer 8

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

Answer 9

yes! we have 0 in voltage change

Answer 10

Yup!

Answer 11

thank you!!

Answer 12

:)

Answer 13

:)