Question

Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacitor if the area of each plate is 5.5 cm2 ? express your answer using two significant figures.

Answer 1

The charge between the plates is  $1.46*10^-^8C$

Data;

• E = 3.0*10^6 V/m
• A = 5.5cm^2 = 5.5*10^-4m^2

The Charge on the Capacitor

The electric field between the plates is

$E = \frac{V}{r}$

• V = potential difference
• r = distance between the plates

The charge on a capacitor is given as

$Q = CV$

• c = capacitor
• v = potential difference
• Q = charge

$C=K\frac{A\epsilon }{d}\\ Q = {KA\epsilon} * E$

Let's substitute the values and solve for the charge on the capacitor

$Q = (5.5*10^-^4*8.85*10^-^1^2)(3.0*10^6)\\Q = 1.46*10^-^8C$

From the calculations above, the charge between the plates is

$1.46*10^-^8C$

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