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blondinia [14]
3 years ago
6

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 capa

citor if the area of each plate is 73 cm^2?
Physics
1 answer:
miskamm [114]3 years ago
7 0

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

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