Two metallic plates `A and B` , each of area `5 xx 10m` are placed parallel to each other at a separation of 1 cm . Plate B carries a positive charge of 33.7 pc . A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0, so that 10 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV . Electric field between the plates at the end of 10 seconds is

Options

A. `2xx10^3 N//C`
B. `10^3 xx N//C`
C. `5 xx10^3 N//C`
D. Zero

Answer

A

Explanation

No. of photoelectron emitted
`=(("No. of photon/unit area/unit time")xx("Area" xx "time"))/10^6`
`=(10^(16) xx5 xx10^(-4) xx10)/10^6=5 xx10^7`
`q_A="ne" =5 xx10^7 xx1.6 xx10^(-19) = 8xx10^(-12)C`
`q_B=33.7 -8 = 25.7pC = 25.7xx10^(-12)C`
Electric field
`E=(qB-q_A)/(2epsilon_0A)= ((25.7-8)xx10^(-12))/(2xx8.85 xx10^(-12)xx5 xx10^(-4))`
`=2 xx10^4 N//C`