A) 2 ´ 103 N/C
B) 103 N/C
C) 5 ´ 103 N/C
D) Zero
Correct Answer: A
Solution :
Number of photoelectrons emitted up to t = 10 sec are \[n=\frac{\begin{align} & (\text{Number of photons per unit area } \\ & \text{ }\,\,\,\,\,\,\text{per unit time)}\times \text{(Area}\times \text{Time)} \\ \end{align}}{\text{1}{{\text{0}}^{\text{6}}}}\] \[=\frac{1}{{{10}^{6}}}[{{(10)}^{16}}\times (5\times {{10}^{-4}})\times (10)]=5\times {{10}^{7}}\] At time t = 10 sec Charge on plate A ; qA = +ne = 5 ´ 107 ´ 1.6 ´ 10?19 = 8 ´ 10?12 C = 8 pC and charge on plate B ; qB = 33.7 ? 8 = 25.7 pc Electric field between the plates \[E=\frac{({{q}_{B}}-{{q}_{A}})}{2\,{{\varepsilon }_{0}}A}=\frac{(25.7-8)\times {{10}^{-12}}}{2\times 8.85\times {{10}^{-12}}\times 5\times {{10}^{-4}}}\]\[=2\times {{10}^{3}}\frac{N}{C}.\]
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