Answer :
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance [tex]C_{0}[/tex] = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed [tex]3.00 \times 10^{4}[/tex] V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = [tex]C \times Ed[/tex]
Therefore, substitute the values into the above formula as follows.
Q = [tex]C \times Ed[/tex]
= [tex]8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})[/tex]
= [tex]2.55 \times 10^{-10} C[/tex]
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is [tex]2.55 \times 10^{-10} C[/tex].