Question

Dry air will break down if the electric field exceeds 3.0*10^6 V/m. What amount of charge can be placed on a parallel-plate capacitor if the area of each plate is 73 cm^2?

Answer 1

Answer:

The charge on each plate of the capacitor is $19.38 \mu C$

with one plate positive and one negative, i.e., $\pm 19.38 \mu C$

Solution:

According to the question:

Critical value of Electric field, $E_{c} = 3.0\times 10^{6} V/m$

Area of each plate of capacitor, $A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}$

Now, the amount of charge on the capacitor's plates can be calculated as:

Capacitance, C  = $\frac{epsilon_{o}\times Area}{Distance, D}$        (1)

Also, Capacitance, C = $\frac{charge, q}{Voltage, V}$

And

Electric field, E = $\frac{Voltage, V}{D}$

So, from the above relations, we can write the eqn for charge, q as:

q = $\epsilon_{o}\times E_{c}\times A_{p}$

q = $8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}$

$q = 19.38 \mu C$