Question

A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N/C. What is the change in voltage? (µC = 1.0 × 10^-6 C)

+1.0 V
-100 V
0 V
+0.20 mV

Answer 1

you could use the equation V=E times R
or Voltage = Electric field times distance

Answer 2

@suhoping

Answer 3

@Nali so would it be 500.0 x 0.2 ?

Answer 4

yes

Answer 5

that gives me 100

Answer 6

just remember to change the cm to m

Answer 7

oh okay so the answer is +1.0 V

Answer 8

yes

Answer 9

thank you so much!

Answer 10

no problem!