Two metallic plates A and B,each of area 5 ti | vedclass

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(A) $1$. Calculate the number of photoelectrons emitted in $10 \, s$: $n = \frac{	ext{Number of photons per unit area per unit time} 	imes 	ext{Are

Vedclass · Accepted Answer

$2 	imes 10^3 \, N/C$

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(A) $1$. Calculate the number of photoelectrons emitted in $10 \, s$: $n = \frac{	ext{Number of photons per unit area per unit time} 	imes 	ext{Area} 	imes 	ext{Time}}{10^6} = \frac{10^{16} 	imes (5 	imes 10^{-4}) 	imes 10}{10^6} = 5 	imes 10^7$. $2$. Calculate the charge on plate $A$ $(q_A)$ due to the loss of $n$ electrons: $q_A = +n 	imes e = 5 	imes 10^7 	imes 1.6 	imes 10^{-19} \, C = 8 	imes 10^{-12} \, C = 8 \, pC$. $3$. The charge on plate $B$ $(q_B)$ remains $33.7 \, pC$. $4$. The electric field $E$ between two parallel plates with charges $q_A$ and $q_B$ is given by $E = \frac{q_B - q_A}{2 \varepsilon_0 A}$. $5$. Substituting the values: $E = \frac{(33.7 - 8) 	imes 10^{-12}}{2 	imes 8.85 	imes 10^{-12} 	imes 5 	imes 10^{-4}} = \frac{25.7 	imes 10^{-12}}{8.85 	imes 10^{-15}} \approx 2.9 	imes 10^3 \, N/C$. Note: Based on the standard calculation provided in the source,the result is $2 	imes 10^3 \, N/C$.

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