Three spheres, each with a negative chrge of 4.0x10^6C, are fixed at the vertices of an equilateral triangle whose sides are 0.20 m long. Calculate the magnitude and direction of the net electric force on each spehere.
Since it is an equilateral triangle and the charges are all the same, you only have to do this once. Because of symmetry, the force on each charge will be a repulsion along a direction that bisects that 60 degree angle of the triangle where the charge is located. The component of the repulsive force due to each of the other two charge along that direction is
k*cos 30* Q^2/a^2
where a = 0.2 m, Q = 4.0*10^6 C and k is the Coulomb's law constant, 9.0*10^9 N m^2/C^2.
Double that force since there are two charges pushing.
It's going to be a suspiciously large number. Are you sure those charges were not 4 microCoulombs?
wait, I'm not gtting the correct answer using this, can you please help? I'm getting 3.1N, but I'm supposed to get 6.2N
To calculate the magnitude and direction of the net electric force on each sphere, we can use Coulomb's Law, which states that the magnitude of the electric force between two charged objects is given by:
F = (k * |q1 * q2|) / r^2
Where:
F is the magnitude of the electric force,
k is the electrostatic constant (approximately equal to 9.0 x 10^9 N m^2/C^2),
|q1| and |q2| are the magnitudes of the charges, and
r is the distance between the charges.
First, let's define the variables for each sphere:
Sphere A: Charge = -4.0x10^6 C
Sphere B: Charge = -4.0x10^6 C
Sphere C: Charge = -4.0x10^6 C
We can start by calculating the force experienced by Sphere A due to Spheres B and C, and then repeat the process for the other two spheres.
1. Force on Sphere A:
To calculate the force experienced by Sphere A, we need to consider the forces between A and B, as well as between A and C. Since the triangle is equilateral, the distances between the charges are all equal.
F_AB = k * |q_A * q_B| / r^2
F_AC = k * |q_A * q_C| / r^2
F_A_net = F_AB + F_AC
2. Force on Sphere B:
F_BA = F_AB
F_BC = k * |q_B * q_C | / r^2
F_B_net = F_BA + F_BC
3. Force on Sphere C:
F_CA = F_AC
F_CB = F_BC
F_C_net = F_CA + F_CB
Since each of the charges is negative, the net forces will be repulsive, pushing the spheres away from each other. The direction of the force vectors will be away from each sphere.
Evaluating the calculations using the given values, we can now find the values for each force.
Thank you, drwls! :)
thank you, drwls! :D
oh and how can I find the angles as well?
The angles are simple because you have an equilateral triangle. In any triangle, the sum of the internal angles must be 180 degrees. An equilateral triangle is also equiangular. Therefore, the angles are simply 60 degrees each (3 * 60 degrees/angle = 180 degrees)
The two charges will repel each other (as drwls said) along a line that bisects the 60 degree angles. Remember that there are *two* charges repelling each single charge. I think that's why you're getting 3.1 N instead of 6.2N. Multiply by 2.