Answer to Question #149501 in Electricity and Magnetism for Nwose Sammy

Question

The plates of parallel capacitor 5.0×10-³m apart maintained at a potential difference of 5.0×10^4. Calculate the magnitude of the a. Electric field intensity between the plates b. Force of the electron c. Acceleration if the electron ( electron charge =1.60×10^-19c)

Answer 1

a) Potential between plates is defined as $U=Ed$ , where $U$-potential between plates, $E$-electric field, $d$-distance between plates. Thus, $E=\dfrac{U}{d}=\dfrac{5\cdot 10^4}{5\cdot 10^{-3}}=10^{7} ~V/m$

b) Force on electron $F=qE\approx1.6\cdot10^{-19}\cdot10^7\approx1.6\cdot 10^{-12}$$N$

c) $F=ma$

$a=\frac{F}{m_{e}}= \frac{1.6\times 10^{-12}}{9.11\times 10^{-31}}= 1.7563\times 10^{18}m/s^{2}$