Answered

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

difference. An electron is located midway between the plates. The magnitude of the electrostatic

force on the electron is 3.8 × 10−16 newton.


Calculate the magnitude of the electric field strength between the plates, in newtons per coulomb

Answer :

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

Cricetus

The magnitude of the electric field strength between the plates will be "2.4 × 10⁻³ V/m".

Electric field

According to the question,

Potential difference, V = 12 V

Electric charge's force, f = 3.8 × 10⁻¹⁶ N

Distance between plates, d = 5.0 × 10⁻³ m

We know the relation,

→ V = E × d

or,

The strength of electric field:

→ E = [tex]\frac{d}{V}[/tex]

By substituting the values,

     = [tex]\frac{12}{5.0\times 10^{-3}}[/tex]

     = 2.4 × 10⁻³ V/m    

Thus the above answer is appropriate.

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