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

• Physics2341313

## Homework Statement

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

## Homework Equations

$F = k q_1*q_2/r^2$

## The Attempt at a Solution

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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 $\hat{r} = 1\sqrt 3 <1, 1, 1>$. 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 $\hat{r}$ notation as it is typically left out in most elementary problems.

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So the force will be in the direction of the unit vector $\hat{r} = 1\sqrt 3 <1, 1, 1>$.
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?

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 $1/\sqrt 3$. So, the result would be

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

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 $F = \sqrt (F_x^2 + F_y^2 + F_z^2)$ but since $F_x$ and $F_z$ are zero the magnitude is just |F_y|

Is this correct?

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