Fig 1. shows an arrangement of three charges 𝑞ଵ = +2𝜇𝐶 𝑞ଶ = −4𝜇𝐶 and q3= +5𝜇C,
placed on the vertices of an equilateral triangle of side 2.0 m.
1.1 Calculate the magnitude and direction of net electrostatic force acting on charge
𝑞ଵ.
To calculate the magnitude and direction of the net electrostatic force acting on charge q1, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electrostatic force between two charges is given by the equation:
F = k * |q1 * q2| / r^2
Where:
- F is the magnitude of the electrostatic force
- k is the electrostatic constant, approximately equal to 9 x 10^9 Nm^2/C^2
- q1 and q2 are the magnitudes of the two charges
- r is the distance between the charges
Step 1: Calculate the distance between the charges:
Since the charges are arranged in an equilateral triangle of side 2.0 m, each side of the triangle is also 2.0 m. The distance between q1 and q2 is the same as the length of one side of the triangle, which is 2.0 m.
Step 2: Calculate the force between q1 and q2:
Using Coulomb's Law, we can calculate the force between q1 and q2:
F12 = k * |q1 * q2| / r^2
= (9 x 10^9 Nm^2/C^2) * |(2 x 10^-6 C) * (4 x 10^-6 C)| / (2.0 m)^2
Simplifying the expression:
F12 = (9 x 10^9 Nm^2/C^2) * (8 x 10^-12 C^2) / 4
= 1.8 x 10^-2 N
The direction of the force between q1 and q2 is attractive (since one charge is positive and the other is negative).
Step 3: Calculate the force between q1 and q3:
Since q3 is also on one of the vertices of the equilateral triangle, the distance between q1 and q3 is also 2.0 m.
Using Coulomb's Law, we can calculate the force between q1 and q3:
F13 = k * |q1 * q3| / r^2
= (9 x 10^9 Nm^2/C^2) * |(2 x 10^-6 C) * (5 x 10^-6 C)| / (2.0 m)^2
Simplifying the expression:
F13 = (9 x 10^9 Nm^2/C^2) * (10 x 10^-12 C^2) / 4
= 2.25 x 10^-2 N
The direction of the force between q1 and q3 is repulsive (since both charges are positive).
Step 4: Calculate the net force on q1:
To calculate the net force on q1, we need to add the vector forces F12 and F13 together. Since the forces are in different directions, we need to consider their magnitudes and their angles.
The magnitude of the net force is given by:
|Fnet| = √(Fx^2 + Fy^2)
Where Fx and Fy are the x and y components of the individual force vectors.
For the x-component:
Fx = F13 - F12
For the y-component:
Fy = 0 (since the force vectors are in the x-y plane and there is no vertical component)
Substituting the values:
|Fnet| = √((F13 - F12)^2 + 0)
= √((2.25 x 10^-2 N - 1.8 x 10^-2 N)^2)
= √(0.45 x 10^-2 N^2)
= 0.67 x 10^-2 N
The direction of the net force can be determined by calculating the angle it makes with the x-axis. Since F13 is larger than F12, the net force will be directed towards q3 (opposite to the direction of F12). Since the x-component of the net force is negative and the y-component is zero, the angle θ can be determined as:
θ = arctan(Fy/Fx)
= arctan(0 / (F13 - F12))
= arctan(0)
The angle is zero degrees, which means the net force is directed along the negative x-axis.
Therefore, the magnitude of the net electrostatic force acting on charge q1 is 0.67 x 10^-2 N, directed along the negative x-axis.