Question
Calculate the distance between 7mf and 7.2mf which have a force of 8N. Take 1÷4piEo=9×10^9
Answers
GPT-4o mini
To calculate the distance between two point charges (which in this case could be described by their capacitance of 7 mf and 7.2 mf respectively) exerting a force of \(8 \, \text{N}\) on each other, we can use Coulomb's Law.
The formula for the electrostatic force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by:
\[
F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}
\]
Where:
- \(F\) is the force between the charges (in newtons),
- \(q_1\) and \(q_2\) are the charges (in coulombs),
- \(r\) is the distance between the charges (in meters),
- \(\epsilon_0\) is the permittivity of free space (approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)).
In this case, since we have capacitance values rather than charge values. The capacitance \(C\) relates to charge \(q\) and voltage \(V\) by:
\[
C = \frac{q}{V}
\]
However, without a voltage value, we can't directly compute the charges from the capacitance values. Instead, we can approach it by using the effective charge values computed from the capacitance (for illustrative purposes) with a defined voltage.
Let's assume \(V\) is the voltage across the capacitance and we'll calculate the charges.
For \(7 \, \text{mf} = 7 \times 10^{-3} \, \text{F}\) and \(7.2 \, \text{mf} = 7.2 \times 10^{-3} \, \text{F}\):
1. **Calculate Charges:**
- If \(V\) is applied across these capacitances,
- Let's define \(V = 1\, V\) (as an example),
- The charges \(q_1\) and \(q_2\) can be computed as:
\[
q_1 = C_1 V = 7 \times 10^{-3} \times 1 = 7 \times 10^{-3} \, \text{C}
\]
\[
q_2 = C_2 V = 7.2 \times 10^{-3} \times 1 = 7.2 \times 10^{-3} \, \text{C}
\]
2. **Using Coulomb’s Law:**
Now we can substitute \(F = 8 \, \text{N}\), \(q_1 = 7 \times 10^{-3} \, \text{C}\), and \(q_2 = 7.2 \times 10^{-3} \, \text{C}\) into Coulomb’s law equation:
\[
8 = \frac{(9 \times 10^9)(7 \times 10^{-3})(7.2 \times 10^{-3})}{r^2}
\]
Rearranging to solve for \(r^2\):
\[
r^2 = \frac{(9 \times 10^9)(7 \times 10^{-3})(7.2 \times 10^{-3})}{8}
\]
Calculating the numerator:
\[
9 \times 10^9 \times 7 \times 10^{-3} \times 7.2 \times 10^{-3} = 9 \times 10^9 \times 50.4 \times 10^{-6} = 453600 \text{ (in appropriate units)}
\]
Hence,
\[
r^2 = \frac{453600}{8} = 56700
\]
Therefore,
\[
r = \sqrt{56700}
\approx 238.1 \text{ meters}
\]
Thus, the calculated distance \(r\) between the two capacitances that exert a force of \(8 \, N\) is approximately \(238.1\) meters.
The formula for the electrostatic force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by:
\[
F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}
\]
Where:
- \(F\) is the force between the charges (in newtons),
- \(q_1\) and \(q_2\) are the charges (in coulombs),
- \(r\) is the distance between the charges (in meters),
- \(\epsilon_0\) is the permittivity of free space (approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)).
In this case, since we have capacitance values rather than charge values. The capacitance \(C\) relates to charge \(q\) and voltage \(V\) by:
\[
C = \frac{q}{V}
\]
However, without a voltage value, we can't directly compute the charges from the capacitance values. Instead, we can approach it by using the effective charge values computed from the capacitance (for illustrative purposes) with a defined voltage.
Let's assume \(V\) is the voltage across the capacitance and we'll calculate the charges.
For \(7 \, \text{mf} = 7 \times 10^{-3} \, \text{F}\) and \(7.2 \, \text{mf} = 7.2 \times 10^{-3} \, \text{F}\):
1. **Calculate Charges:**
- If \(V\) is applied across these capacitances,
- Let's define \(V = 1\, V\) (as an example),
- The charges \(q_1\) and \(q_2\) can be computed as:
\[
q_1 = C_1 V = 7 \times 10^{-3} \times 1 = 7 \times 10^{-3} \, \text{C}
\]
\[
q_2 = C_2 V = 7.2 \times 10^{-3} \times 1 = 7.2 \times 10^{-3} \, \text{C}
\]
2. **Using Coulomb’s Law:**
Now we can substitute \(F = 8 \, \text{N}\), \(q_1 = 7 \times 10^{-3} \, \text{C}\), and \(q_2 = 7.2 \times 10^{-3} \, \text{C}\) into Coulomb’s law equation:
\[
8 = \frac{(9 \times 10^9)(7 \times 10^{-3})(7.2 \times 10^{-3})}{r^2}
\]
Rearranging to solve for \(r^2\):
\[
r^2 = \frac{(9 \times 10^9)(7 \times 10^{-3})(7.2 \times 10^{-3})}{8}
\]
Calculating the numerator:
\[
9 \times 10^9 \times 7 \times 10^{-3} \times 7.2 \times 10^{-3} = 9 \times 10^9 \times 50.4 \times 10^{-6} = 453600 \text{ (in appropriate units)}
\]
Hence,
\[
r^2 = \frac{453600}{8} = 56700
\]
Therefore,
\[
r = \sqrt{56700}
\approx 238.1 \text{ meters}
\]
Thus, the calculated distance \(r\) between the two capacitances that exert a force of \(8 \, N\) is approximately \(238.1\) meters.