Question

the plates of a parallel plate capacitor,5.0x10^-3m apart are maintained at a potential difference of 5.0x10^4 calculate the magnitude of the i. electric field intensity between the plates ii. force on the electron.

Answer 1

For a parallel plate capacitor, $E=V/d$

$\implies E=5*10^4/5*10^{-3}=10^7Vm^{-1}$

$Force =qE=1.6*10^{-19}*10^7=1.6*10^{-12} N$