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 field has a strength of 500.0 N/C. What is the change in voltage?

-100 V
+1.0 V
0 V
+0.20 mV

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 field has a strength of 500.0 N/C. What is the change in voltage?

       -100 V
       +1.0 V
       0 V
       +0.20 mV

agent0smith · Answer

@musicfuryu Sorry, that should've been $$\large E = \frac{ \Delta V }{ d }$$

E is electric field strength (in N/C)
d is distance (in metres, make sure you convert your distance from cm to m)
V is the change in voltage, which you need to find. 

Note that the amount of charge does not affect the change in voltage, only the distance travelled affects voltage.

Rearranging gives:
$$\Large E d = \Delta V$$