Answer:
The charge on each plate of the capacitor is [tex]19.38 \mu C[/tex]
with one plate positive and one negative, i.e., [tex]\pm 19.38 \mu C[/tex]
Solution:
According to the question:
Critical value of Electric field, [tex]E_{c} = 3.0\times 10^{6} V/m[/tex]
Area of each plate of capacitor, [tex]A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}[/tex]
Now, the amount of charge on the capacitor's plates can be calculated as:
Capacitance, C = [tex]\frac{epsilon_{o}\times Area}{Distance, D}[/tex] (1)
Also, Capacitance, C = [tex]\frac{charge, q}{Voltage, V}[/tex]
And
Electric field, E = [tex]\frac{Voltage, V}{D}[/tex]
So, from the above relations, we can write the eqn for charge, q as:
q = [tex]\epsilon_{o}\times E_{c}\times A_{p}[/tex]
q = [tex]8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}[/tex]
[tex]q = 19.38 \mu C[/tex]
