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 3.9 \times 10^{-15}N acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates?

# Solution:

### a) To find the electric field, we use E=\frac{F}{e} where e is the elementary charge, or the charge carried by a single proton and F is the electric force. The elementary charge is e=1.60\times10^{-19}C.

### Thus,

### E=\frac{F}{e}= \frac{(3.9\times10^{-15}N)}{(1.60\times10^{-19}C)}

### =2.4\times10^{4}\frac{N}{C}=2.4\times10^{4}\frac{V}{m}

### b) To find the potential difference between the plates, we multiply the electric field value by the distance between the plates.

### \triangle V= E\triangle s =(2.4 \times10^{4}\frac{N}{C})(0.12m)=2.9 \times 10^{3}V

Did you mean 3.9 instead of 3.5

Yes, that was a type. We fixed it 🙂 Many thanks!