Question

Two conducting parallel plates 5.0 × 10−3 meter apart are charged with a 12-volt potential

Answer 1

Answer: 2.4×10^-3 v/m

Explanation: distance between plates of capacitor (d) =5.0×10^-3m

Potential difference between plates (v) = 12v

Force on electronic charge (f) = 3.8×10^-16 N

Strength of electric field (E) =?

The formulae that relates potential difference, eoectiic field strength and distance between plates is given as

v = Ed

By substituting the parameters, we have that

12 = E × 5.0×10^-3

E = 12/ 5.0 × 10^-3

E = 2.4×10^-3 v/m