Question

If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk capacitor of radius 54 cm with a gap of 3 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)? The constant ε0 = 8.85 ✕ 10-12 C2/(N·m2).

Answer 1

The electric field inside a parallel plate capacitor is given by:

E = Q/(ε₀A)

E is the electric field, Q is the charge stored on one of the plates, and A is the area of one of the plates.

The plates are circular, so the area A of one of the plates is given by:

A = πr²

where r is the radius.

Therefore the electric field is given by:

E = Q/(ε₀πr²)

Given values:

E = 3×10⁶N/C (max E field allowed before breakdown occurs)

r = 54×10⁻²m

Plug in these values and solve for Q:

3×10⁶ = Q/(ε₀π(54×10⁻²)²)

Q = 2.4×10⁻⁵C