Question

Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of 15 3.9 10 N − ⋅ acts on an electron if it is placed anywhere between the two plates. (a) Find the electric field magnitude at the position of the electron. (b) What is the potential difference between the plates?

Answer 1

Answer:

a) 2.4×10^4N/C

b) 2.9 ×10^3V

Explanation:

Correct ststement: An electric force of 3.9×10^-15N acts on the electron if it is placed anywhere between the two plates.

a) The electric field magnitude is given by E= F/e

Where F = electric force

e= elementary charge carried by a single proton e= 1.6×10^-19C

E= (3.9×10^-15)/(1.6×10^-19)

E= 2.4×10^4NC

b) Ptential difference is given by:

Change in V= E×change in distance

Potential difference= (2.4×10^4)× (0.12)

Potential difgerence= 2.9×10^3V