two point charges q1=3.5 mC and q2=2.92 pC are located on (2,3)m and (-3,5)m of a Cartesian plane, respectively. Determine the magnitude and nature of the force experience by the two charges.
To determine the magnitude and nature of the force experienced by the two charges, we can use the Coulomb's Law equation:
F = (k * |q1 * q2|) / r^2
Where:
- F is the magnitude of the force between the charges,
- k is Coulomb's constant (k ≈ 9 × 10^9 Nm^2/C^2),
- q1 and q2 are the magnitudes of the charges,
- |q1 * q2| is the product of the magnitudes of the charges (ignoring the signs),
- r is the distance between the charges.
Let's calculate it step by step:
1. Convert the charges to Coulombs:
- q1 = 3.5 mC = 3.5 × 10^-3 C
- q2 = 2.92 pC = 2.92 × 10^-12 C
2. Calculate the distance between the charges:
- Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
- For q1: (x1, y1) = (2, 3)
- For q2: (x2, y2) = (-3, 5)
- Distance = √[(-3 - 2)^2 + (5 - 3)^2]
= √[25 + 4]
= √29
3. Calculate the magnitude of the force:
- F = (k * |q1 * q2|) / r^2
- F = (9 × 10^9 Nm^2/C^2 * |3.5 × 10^-3 C * 2.92 × 10^-12 C|) / (√29)^2
- F = (9 × 10^9 Nm^2/C^2 * 1.022 × 10^-14 C^2) / 29
- F ≈ 3.17 × 10^-4 N
4. Determine the nature of the force:
The nature of the force can be determined by the signs of the charges. If the charges are of the same sign (either positive or negative), the force is repulsive. If the charges are of opposite signs, the force is attractive.
- q1 = 3.5 mC = positive charge
- q2 = 2.92 pC = positive charge
Since both charges are positive, the force between them is repulsive.
Therefore, the magnitude of the force experienced by the two charges is approximately 3.17 × 10^-4 N and the nature of the force is repulsive.