Three charges 𝑞1 = +2𝜇𝐶 𝑞2 = −4𝜇𝐶 and q3=+5𝜇C are placed on the vertices of an equilateral triangle of side 2.0 m.

Question

1.1 Calculate the magnitude and direction of net electrostatic force acting on charge 𝑞1.

Answer 1

To calculate the net electrostatic force acting on charge 𝑞1, we need to find the forces between 𝑞1 and 𝑞2, and between 𝑞1 and 𝑞3. The force between two charges can be calculated using Coulomb's Law:

𝐹 = 𝑘 * (𝑞1 * 𝑞2) / 𝑟^2

Where:
- 𝐹 is the force between the charges
- 𝑘 is Coulomb's constant (9.0 x 10^9 Nm^2/C^2)
- 𝑞1 and 𝑞2 are the magnitudes of the charges
- 𝑟 is the distance between the charges

Let's calculate the force between 𝑞1 and 𝑞2 first. Since the triangle is equilateral, the distance between them is equal to the side length of the triangle, which is 2.0 m.
𝐹12 = (9.0 x 10^9 Nm^2/C^2) * ((2 μC * -4 μC) / (2.0 m)^2)

Simplifying this expression:
𝐹12 = (9.0 x 10^9 Nm^2/C^2) * (-8 𝜇C^2) / 4.0 m^2

𝐹12 = -18𝜇N (Note that the direction is attractive, since the charges are opposite in sign)

Now let's calculate the force between 𝑞1 and 𝑞3. Again, the distance between them is 2.0 m.
𝐹13 = (9.0 x 10^9 Nm^2/C^2) * ((2 μC * 5 μC) / (2.0 m)^2)

Simplifying this expression:
𝐹13 = (9.0 x 10^9 Nm^2/C^2) * (10 𝜇C^2) / 4.0 m^2

𝐹13 = 45𝜇N (The direction is repulsive, since the charges are the same in sign)

To find the net force on 𝑞1, we need to find the vector sum of 𝐹12 and 𝐹13. Since they are along the same line, we can simply add their magnitudes.
𝐹net = |𝐹12| + |𝐹13|

𝐹net = |-18𝜇N| + |45𝜇N|

𝐹net = 63𝜇N

Therefore, the magnitude of the net electrostatic force acting on charge 𝑞1 is 63 𝜇N, and the direction is towards 𝑞3.