Three equal charges are placed at the corners of an equilateral triangle 0.46 m on a side. What are the magnitude of the force on each charge if the charges are each -4.1×10-9 C?
Perform a vector addition of the Coulomb forces. The resultant force on each charge should be a repulsion away from the opposite base, along the angle bisector.
can you explain in detail because i tried to follow along with what my professor taught in class, but i couldn't understand anything.
never mind, i figured it out. thanks
To find the magnitude of the force on each charge, we can use Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:
F = (k * q1 * q2) / r^2
Where:
F is the magnitude of the force,
k is the electrostatic constant (9 x 10^9 Nm^2/C^2),
q1 and q2 are the charges,
and r is the distance between the charges.
In this case, we have an equilateral triangle with three charges at the corners. Each charge is -4.1×10^-9 C.
The distance between the charges is the length of one side of the equilateral triangle, which is 0.46 m.
Using Coulomb's law, we can calculate the magnitude of the force on each charge.
F = (k * q1 * q2) / r^2
F = (9 x 10^9 Nm^2/C^2 * -4.1×10^-9 C * -4.1×10^-9 C) / (0.46 m)^2
Calculating this expression will give us the magnitude of the force on each charge.