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Sagot :
Sure, let's go through the steps to calculate the net force on sphere [tex]\(X\)[/tex] due to the forces from sphere [tex]\(Y\)[/tex] and sphere [tex]\(Z\)[/tex].
Given:
- Charge of sphere [tex]\(X\)[/tex] ([tex]\(q_X\)[/tex]) [tex]\(= 5 \times 10^{-6} \, C\)[/tex]
- Charge of sphere [tex]\(Y\)[/tex] ([tex]\(q_Y\)[/tex]) [tex]\(= 7 \times 10^{-6} \, C\)[/tex]
- Charge of sphere [tex]\(Z\)[/tex] ([tex]\(q_Z\)[/tex]) [tex]\(= -4 \times 10^{-6} \, C\)[/tex]
- Distance between [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] ([tex]\(r_{XY}\)[/tex]) [tex]\(= 0.02 \, m\)[/tex]
- Distance between [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] ([tex]\(r_{XZ}\)[/tex]) [tex]\(= 0.03 \, m\)[/tex]
- Coulomb's constant ([tex]\(k\)[/tex]) [tex]\(= 8.99 \times 10^9 \, N \cdot m^2/C^2\)[/tex]
1. Calculate the force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Y\)[/tex]:
Using Coulomb's law:
[tex]\[ F_{XY} = k \frac{q_X \cdot q_Y}{r_{XY}^2} \][/tex]
Substituting the given values:
[tex]\[ F_{XY} = 8.99 \times 10^9 \frac{(5 \times 10^{-6}) \cdot (7 \times 10^{-6})}{(0.02)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{XY} = 786.625 \, N \][/tex]
2. Calculate the force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Z\)[/tex]:
Using Coulomb's law:
[tex]\[ F_{XZ} = k \frac{q_X \cdot q_Z}{r_{XZ}^2} \][/tex]
Substituting the given values:
[tex]\[ F_{XZ} = 8.99 \times 10^9 \frac{(5 \times 10^{-6}) \cdot (-4 \times 10^{-6})}{(0.03)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{XZ} = -199.778 \, N \][/tex]
Here, the negative sign indicates that the force is attractive (since the charges are of opposite signs).
3. Determine the resultant force on sphere [tex]\(X\)[/tex]:
Since sphere [tex]\(Z\)[/tex] is placed at right angles to sphere [tex]\(X\)[/tex], the forces [tex]\(F_{XY}\)[/tex] and [tex]\(F_{XZ}\)[/tex] are perpendicular to each other. Therefore, we use the Pythagorean theorem to find the magnitude of the net force:
[tex]\[ F_{net} = \sqrt{F_{XY}^2 + F_{XZ}^2} \][/tex]
Substituting the calculated values:
[tex]\[ F_{net} = \sqrt{786.625^2 + (-199.778)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{net} = 811.597 \, N \][/tex]
Thus, the magnitudes of the forces are:
- Force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Y\)[/tex] ([tex]\(F_{XY}\)[/tex]): [tex]\(786.625 \, N\)[/tex]
- Force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Z\)[/tex] ([tex]\(F_{XZ}\)[/tex]): [tex]\(-199.778 \, N\)[/tex] (indicating attraction)
- Net force on sphere [tex]\(X\)[/tex] due to sphere [tex]\(Y\)[/tex] and sphere [tex]\(Z\)[/tex] ([tex]\(F_{net}\)[/tex]): [tex]\(811.597 \, N\)[/tex]
Given:
- Charge of sphere [tex]\(X\)[/tex] ([tex]\(q_X\)[/tex]) [tex]\(= 5 \times 10^{-6} \, C\)[/tex]
- Charge of sphere [tex]\(Y\)[/tex] ([tex]\(q_Y\)[/tex]) [tex]\(= 7 \times 10^{-6} \, C\)[/tex]
- Charge of sphere [tex]\(Z\)[/tex] ([tex]\(q_Z\)[/tex]) [tex]\(= -4 \times 10^{-6} \, C\)[/tex]
- Distance between [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] ([tex]\(r_{XY}\)[/tex]) [tex]\(= 0.02 \, m\)[/tex]
- Distance between [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] ([tex]\(r_{XZ}\)[/tex]) [tex]\(= 0.03 \, m\)[/tex]
- Coulomb's constant ([tex]\(k\)[/tex]) [tex]\(= 8.99 \times 10^9 \, N \cdot m^2/C^2\)[/tex]
1. Calculate the force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Y\)[/tex]:
Using Coulomb's law:
[tex]\[ F_{XY} = k \frac{q_X \cdot q_Y}{r_{XY}^2} \][/tex]
Substituting the given values:
[tex]\[ F_{XY} = 8.99 \times 10^9 \frac{(5 \times 10^{-6}) \cdot (7 \times 10^{-6})}{(0.02)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{XY} = 786.625 \, N \][/tex]
2. Calculate the force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Z\)[/tex]:
Using Coulomb's law:
[tex]\[ F_{XZ} = k \frac{q_X \cdot q_Z}{r_{XZ}^2} \][/tex]
Substituting the given values:
[tex]\[ F_{XZ} = 8.99 \times 10^9 \frac{(5 \times 10^{-6}) \cdot (-4 \times 10^{-6})}{(0.03)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{XZ} = -199.778 \, N \][/tex]
Here, the negative sign indicates that the force is attractive (since the charges are of opposite signs).
3. Determine the resultant force on sphere [tex]\(X\)[/tex]:
Since sphere [tex]\(Z\)[/tex] is placed at right angles to sphere [tex]\(X\)[/tex], the forces [tex]\(F_{XY}\)[/tex] and [tex]\(F_{XZ}\)[/tex] are perpendicular to each other. Therefore, we use the Pythagorean theorem to find the magnitude of the net force:
[tex]\[ F_{net} = \sqrt{F_{XY}^2 + F_{XZ}^2} \][/tex]
Substituting the calculated values:
[tex]\[ F_{net} = \sqrt{786.625^2 + (-199.778)^2} \][/tex]
Upon evaluation:
[tex]\[ F_{net} = 811.597 \, N \][/tex]
Thus, the magnitudes of the forces are:
- Force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Y\)[/tex] ([tex]\(F_{XY}\)[/tex]): [tex]\(786.625 \, N\)[/tex]
- Force between sphere [tex]\(X\)[/tex] and sphere [tex]\(Z\)[/tex] ([tex]\(F_{XZ}\)[/tex]): [tex]\(-199.778 \, N\)[/tex] (indicating attraction)
- Net force on sphere [tex]\(X\)[/tex] due to sphere [tex]\(Y\)[/tex] and sphere [tex]\(Z\)[/tex] ([tex]\(F_{net}\)[/tex]): [tex]\(811.597 \, N\)[/tex]
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