Force at center of cube due to four charges of equal mag.

In summary, the problem involves four charges of 2*10^-7 placed on the corners of one face of a cube of 15 cm and a charge of -2 * 10^-7 C placed at the center of the cube. The force on the charge at the center of the cube is found using the equation F = k q_1*q_2/r^2 and considering the direction of the unit vector \hat{r} = 1\sqrt 3 <1, 1, 1>. The magnitude of the vector is found by normalizing it with a prefactor of 1/\sqrt 3. The resulting magnitude is |F_y|, since the other forces in the x and z directions have
  • #1
Physics2341313
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Homework Statement



Four charges of [itex]2*10^-7[/itex] are placed on the corners of one face of a cube of 15 cm. A charge of [itex]-2 * 10^-7[/itex] C is placed at the center of the cube. What is the force on the charge at the center of the cube?

Homework Equations



[itex] F = k q_1*q_2/r^2 [/itex]

The Attempt at a Solution


[/B]
I understand how to get the correct answer. Depending upon where we place the charges on one of the cubes faces the forces will be in the direction normal to that face. (sides = y-components, back-faces = x, bottom/top = z-components) and everything else will cancel. So the force will be in the direction of the unit vector [itex]\hat{r} = 1\sqrt 3 <1, 1, 1>[/itex]. This is the part I don't understand, the unit vector's magnitude must be one so makes sense that 3(1/3) = 1 trivially. But, how is the unit vector here giving the force in said direction(s)? Having some trouble with the [itex]\hat{r}[/itex] notation as it is typically left out in most elementary problems.
 
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  • #2
Physics2341313 said:
So the force will be in the direction of the unit vector [itex]\hat{r} = 1\sqrt 3 <1, 1, 1>[/itex].
The force coming from one specific charge? This vector is along a volume diagonal of the cube, so it is not parallel to the total force.

What is the magnitude of the vector (1,1,1)? To normalize the vector, which prefactor do you need?
 
  • #3
I think I messed up a little bit. The diagonal will be the side length, a, times square root of three.
I only need one face so I just need to multiply the result by [itex]1/\sqrt 3[/itex]. So, the result would be

[itex] F_Q = 4[1/4\pi \epsilon_o * q^2/r^2 ] * 1\sqrt 3 [/itex]

In the way I worked it I placed the charges one the left-most side of the cube so the x and z components of the force on the middle charge have a net contribution of 0 leaving the force to be in the y direction. Since the other forces contribute zero along their respective direction the magnitude would be [itex] F = \sqrt (F_x^2 + F_y^2 + F_z^2) [/itex] but since [itex]F_x[/itex] and [itex]F_z[/itex] are zero the magnitude is just |F_y|

Is this correct?
 
  • #4
Correct, but don't forget brackets in the denominator. Those are not necessary on paper, but here they are. Alternatively, use LaTeX:
$$\frac{1}{4 \pi \epsilon_0}$$
 

Related to Force at center of cube due to four charges of equal mag.

What is the force at the center of a cube due to four charges of equal magnitude?

The force at the center of a cube due to four charges of equal magnitude is zero. This is because the forces from each charge cancel out due to the symmetry of the cube.

How do you calculate the force at the center of a cube due to four charges?

To calculate the force at the center of a cube due to four charges, you can use Coulomb's Law. First, calculate the distance between the center of the cube and each charge. Then, calculate the force between the center and each charge using Coulomb's Law. Finally, add up all the individual forces to find the total force at the center.

Why is the force at the center of a cube due to four charges zero if the charges are not equal in magnitude?

The force at the center of a cube due to four charges will only be zero if the charges are equal in magnitude. If the charges are not equal, the forces will not cancel out and there will be a resulting force at the center of the cube.

Can the force at the center of a cube due to four charges ever be non-zero?

Yes, the force at the center of a cube due to four charges can be non-zero if the charges are not equal in magnitude. In this case, the forces will not cancel out and there will be a resulting force at the center of the cube.

What is the relationship between the distance between the charges and the force at the center of the cube?

The force at the center of a cube due to four charges is inversely proportional to the distance between the charges. This means that as the distance between the charges increases, the force at the center decreases, and vice versa.

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