Three charges q, 2q, and 3q are arragend to form a equileteral triangle. A charge Q is placed at the center of the triagle. What´s the Net Force about Q. Give the modulus and the direcction of the force.
The forces are 120 degrees apart. Lets let the q be the reference charge for angles, measured from the center, and postive angles toward 2q is at 120, and 3q is at 240.
Force=kq/s^2 (q at 0 + 2q at 120 + 3q at 240)
now, but the magic of superposition, we can take out q from each corner, because the resultant force from those three equally spaced charges is zero)
Force=kq/s^2 (q at 120 + 2 q at 240)
=kq^2/s^2 (1 @ 120 + 2@240)
= kq^2/s^2 sqrt5 is the magnitude
direction(1sin120@180+1cos120@180+2sin240@90+2 cos240 @90)
combine like terms in the ( ) YOu have now components which are 90 degrees apart, which can be added to find the direction. Draw a sketch of this, you can see it visually.
To find the net force on charge Q at the center of the equilateral triangle, we need to calculate the individual forces exerted by each charge and add them vectorially.
Let's assume that q, 2q, and 3q are positive charges. Since all charges are equidistant from charge Q, we can assume that they lie along the sides of an equilateral triangle.
The force between two charges is given by Coulomb's Law:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the Coulomb's constant (9 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
First, let's calculate the forces exerted on charge Q by each of the other charges. Since they are in equilibrium (forming an equilateral triangle), the forces must cancel out.
1. Force exerted on charge Q by charge q (on the left side):
F1 = (k * q * Q) / r^2
2. Force exerted on charge Q by charge 2q (on the right side):
F2 = (k * 2q * Q) / r^2
3. Force exerted on charge Q by charge 3q (at the top):
F3 = (k * 3q * Q) / r^2
Since the charges are symmetrically arranged, F2 will equal F3 in magnitude but have opposite directions. The magnitudes of both F2 and F3 can be calculated as follows:
F2 = F3 = (k * 2q * Q) / r^2
Since the forces F2 and F3 are opposite in direction, their vector sum is equal to zero.
Therefore, the net force on charge Q is the sum of F1, F2, and F3:
Net Force = F1
To determine the modulus and direction of the force, we'll need to consider the angles involved. Let's call the angles of the equilateral triangle ∠1, ∠2, ∠3.
The direction of the force can be found by drawing a vector triangle with the forces F1, F2, and F3 as sides.
Since F2 and F3 cancel out each other (as described before), the net force is directed along the line joining charge Q and charge q. Let's call this line the x-axis.
Therefore, the modulus of the force is given by:
|Net Force| = |F1| = (k * q * Q) / r^2
As for the direction, the net force will act along the positive x-axis since it is the only force present in that direction.
So, to summarize:
- The modulus of the net force on charge Q is (k * q * Q) / r^2.
- The direction of the net force is along the positive x-axis.