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To determine the number of coulombs of charge needed to create a spark in the air, given an electric field and a separation distance, we can utilize the laws of electrostatics and the properties of a parallel plate capacitor.
Here are the known values:
- Electric field ([tex]\(E\)[/tex]) = [tex]\(3 \times 10^{16} \frac{N}{C}\)[/tex]
- Distance ([tex]\(d\)[/tex]) = 1 mm = [tex]\(1 \times 10^{-3}\)[/tex] meters
We will use the relation for the electric field in a capacitor:
[tex]\[ E = \frac{V}{d} \][/tex]
where [tex]\(V\)[/tex] is the potential difference and [tex]\(d\)[/tex] is the separation distance. From this, we can infer that:
[tex]\[ V = E \times d \][/tex]
Next, using the formula for charge ([tex]\(Q\)[/tex]) on a capacitor:
[tex]\[ Q = C \times V \][/tex]
where [tex]\(C\)[/tex] is the capacitance and [tex]\(V\)[/tex] is the potential difference.
The capacitance ([tex]\(C\)[/tex]) of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\epsilon_0 \times A}{d} \][/tex]
where:
- [tex]\(\epsilon_0\)[/tex] is the permittivity of free space ([tex]\(8.854 \times 10^{-12} \frac{F}{m}\)[/tex])
- [tex]\(A\)[/tex] is the area of the plates (which cancels out as it is not provided and doesn't affect our computation in this case)
Substituting the formula for [tex]\(C\)[/tex] into the expression for [tex]\(Q\)[/tex], we obtain:
[tex]\[ Q = \left(\frac{\epsilon_0 \times A}{d}\right) \times V \][/tex]
Since [tex]\(V = E \times d\)[/tex], subbing [tex]\(V\)[/tex] into the equation gives:
[tex]\[ Q = \left(\frac{\epsilon_0 \times A}{d}\right) \times (E \times d) \][/tex]
Simplifying this, the [tex]\(d\)[/tex] terms cancel out, and we get:
[tex]\[ Q = \epsilon_0 \times A \times E \][/tex]
However, since the area [tex]\(A\)[/tex] is not specified, we conclude:
[tex]\[ Q = \epsilon_0 \times E \times d \][/tex]
Substitute the known values:
[tex]\[ \epsilon_0 = 8.854 \times 10^{-12} \frac{F}{m} \][/tex]
[tex]\[ E = 3 \times 10^{16} \frac{N}{C} \][/tex]
[tex]\[ d = 1 \times 10^{-3} \text{ meters} \][/tex]
[tex]\[ Q = (8.854 \times 10^{-12}) \times (3 \times 10^{16}) \times (1 \times 10^{-3}) \][/tex]
Carrying out the multiplication step-by-step:
[tex]\[ Q = 8.854 \times 3 = 26.562 \][/tex]
[tex]\[ 26.562 \times 10^{-12 + 16 - 3} = 26.562 \times 10^1 = 265.62 \][/tex]
Rounding this to the nearest whole number, we get:
[tex]\[ Q \approx 266 \text{ Coulombs} \][/tex]
Hence, the number of coulombs of charge needed for a spark is approximately:
[tex]\[ 266 \text{ C} \][/tex]
Here are the known values:
- Electric field ([tex]\(E\)[/tex]) = [tex]\(3 \times 10^{16} \frac{N}{C}\)[/tex]
- Distance ([tex]\(d\)[/tex]) = 1 mm = [tex]\(1 \times 10^{-3}\)[/tex] meters
We will use the relation for the electric field in a capacitor:
[tex]\[ E = \frac{V}{d} \][/tex]
where [tex]\(V\)[/tex] is the potential difference and [tex]\(d\)[/tex] is the separation distance. From this, we can infer that:
[tex]\[ V = E \times d \][/tex]
Next, using the formula for charge ([tex]\(Q\)[/tex]) on a capacitor:
[tex]\[ Q = C \times V \][/tex]
where [tex]\(C\)[/tex] is the capacitance and [tex]\(V\)[/tex] is the potential difference.
The capacitance ([tex]\(C\)[/tex]) of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\epsilon_0 \times A}{d} \][/tex]
where:
- [tex]\(\epsilon_0\)[/tex] is the permittivity of free space ([tex]\(8.854 \times 10^{-12} \frac{F}{m}\)[/tex])
- [tex]\(A\)[/tex] is the area of the plates (which cancels out as it is not provided and doesn't affect our computation in this case)
Substituting the formula for [tex]\(C\)[/tex] into the expression for [tex]\(Q\)[/tex], we obtain:
[tex]\[ Q = \left(\frac{\epsilon_0 \times A}{d}\right) \times V \][/tex]
Since [tex]\(V = E \times d\)[/tex], subbing [tex]\(V\)[/tex] into the equation gives:
[tex]\[ Q = \left(\frac{\epsilon_0 \times A}{d}\right) \times (E \times d) \][/tex]
Simplifying this, the [tex]\(d\)[/tex] terms cancel out, and we get:
[tex]\[ Q = \epsilon_0 \times A \times E \][/tex]
However, since the area [tex]\(A\)[/tex] is not specified, we conclude:
[tex]\[ Q = \epsilon_0 \times E \times d \][/tex]
Substitute the known values:
[tex]\[ \epsilon_0 = 8.854 \times 10^{-12} \frac{F}{m} \][/tex]
[tex]\[ E = 3 \times 10^{16} \frac{N}{C} \][/tex]
[tex]\[ d = 1 \times 10^{-3} \text{ meters} \][/tex]
[tex]\[ Q = (8.854 \times 10^{-12}) \times (3 \times 10^{16}) \times (1 \times 10^{-3}) \][/tex]
Carrying out the multiplication step-by-step:
[tex]\[ Q = 8.854 \times 3 = 26.562 \][/tex]
[tex]\[ 26.562 \times 10^{-12 + 16 - 3} = 26.562 \times 10^1 = 265.62 \][/tex]
Rounding this to the nearest whole number, we get:
[tex]\[ Q \approx 266 \text{ Coulombs} \][/tex]
Hence, the number of coulombs of charge needed for a spark is approximately:
[tex]\[ 266 \text{ C} \][/tex]
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