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(C) In the space charge limited region,the plate current $(i_p)$ is governed by Child's law,which states that $i_p = K V_p^{3/2}$,where $K$ is a const

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$80$

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(C) In the space charge limited region,the plate current $(i_p)$ is governed by Child's law,which states that $i_p = K V_p^{3/2}$,where $K$ is a constant and $V_p$ is the plate voltage.Given: $i_{p1} = 10\, mA$ at $V_{p1} = 150\, V$.We need to find $i_{p2}$ at $V_{p2} = 600\, V$.Using the ratio: $\frac{i_{p2}}{i_{p1}} = \left( \frac{V_{p2}}{V_{p1}} ight)^{3/2}$.Substituting the values: $\frac{i_{p2}}{10} = \left( \frac{600}{150} ight)^{3/2} = (4)^{3/2}$.Calculating the power: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$i_{p2} = 10 	imes 8 = 80\, mA$.

In the space charge limited region,the plate current in a diode is $10\, mA$ for a plate voltage of $150\, V$. If the plate voltage is increased to $600\, V$,then the plate current will be.......$mA$.

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