Question: With the potential difference between two parallel conducting plates i

Question

Question:

With the potential difference between two parallel conducting plates is 16.0 V. The area on each plate is 4.8cm^2 and the spacing between the plates is 0.12 mm.

Find the electric field between the two plates, the electric forceon an electron between the two plates, the work done by the electric field in moving the electron from the negatively charged plate to the positively charged plate.

*my answer for the Electric field is: 1.33x10^5 N/V not sure if correct.

If can please give details of the solution.

Explanation / Answer

a)

E=V/d =16/(0.12*10^-3)

E=1.33*10^5 N/C

b)

Capacitance

C=eoA/d =(8.85*10^-12)*(4.8*10^-4)/(0.12*10^-3)

C=3.54*10^-11 F

charge

Q=CV =16*3.54*10^-11

Q=5.664*10^-10 C

force on an electron

F=QE =5.664*10^-10*1.33*10^5

F=7.55*10^-5 N ot 75.5 uN

c)

Work done

W=(1/2)CV^2 =(1/2)*(3.54*10^-11)*16^2

W=4.53*10^-9 J or 4.53 nJ

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