A potential difference of 4.75 kV is established between parallel plates in airf the air becomes ionized (and hence electrically conducting) when the electric field exceeds 3.00×106 V/m, what is the minimum separation the plates can have without ionizing the air?

Question

A potential difference of 4.75 kV is established between parallel plates in airf the air becomes ionized (and hence electrically conducting) when the electric field exceeds 3.00×106 V/m, what is the minimum separation the plates can have without ionizing the air?

d=___________m

Answer 1

V/d(min) = 3*10^6 V/m, the breakdown potential

d(min) = (4.75*10^3 V)/(3*10^6 V/m)
= 1.58*10^-3 m
= 1.58 mm

Answer 2

A potential difference of 10 kV is established between parallel plates in air.

(a) If the air becomes electrically conducting when the electric field exceeds
4*106 V/m, what is the minimum separation of the plates?
(b) When the separation has the minimum value calculated in part (a), what
is the surface charge density on each plate?

Answer 3

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Answer 4

Well, you know, with all that potential difference, you'd think the air would start feeling a bit "charged," huh? Anyway, let's calculate the minimum separation, aka the "distance" between the plates.

We can use the formula: electric field strength (E) = potential difference (V) / separation distance (d). So, rearranging the formula, we get:

d = V / E

Substituting the given values, we have:

d = 4,750 V / 3,000,000 V/m

Dividing those numbers, we find:

d ≈ 0.001583 m

So the minimum separation between the plates is approximately 0.001583 meters, or about 1.58 millimeters. Just enough for a teensy bit of electrical humor!

Answer 5

To determine the minimum separation between the plates (d) without ionizing the air, we need to find the electric field generated by the potential difference and compare it with the threshold electric field for ionization.

The electric field (E) between parallel plates can be calculated using the formula:

E = V / d

Where:
E is the electric field (in V/m)
V is the potential difference (in volts)
d is the separation between the plates (in meters)

Given:
V = 4.75 kV = 4.75 × 10^3 V
Threshold Electric Field (E_threshold) = 3.00 × 10^6 V/m

We can rearrange the formula to solve for the separation (d):

d = V / E

Substituting the values:

d = (4.75 × 10^3 V) / (3.00 × 10^6 V/m)

Now, let's calculate:

d = 4.75 × 10^3 / 3.00 × 10^6 = 1.5833... × 10^-3 m

Rounding to the appropriate number of significant figures:

d ≈ 1.58 × 10^-3 m

Therefore, the minimum separation between the plates (d) without ionizing the air is approximately 1.58 mm (or 1.58 × 10^-3 m).