Point charge, proton placement, x-, y-axis

[kuzco890]

Homework Statement

A point charge q = -0.368 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on it to be exactly opposite to its weight? Let the y-axis be vertical (positive up and negative down) and the x-axis be horizontal (positive to the right and negative to the left. Enter the x-value first and then the y-value.

Homework Equations

mass of proton = 1.67e-27 kg
charge of proton = 1.60e-19 Coulombs

The Attempt at a Solution

electric force = weight?
[(8.9e9 Nm^2/C^2)(1.6e-19 C)(-0.368e-9 C)]/[r^2] = (1.67e-27 kg)(9.80 m/s)
solve for r. r = 5.66e3

But I don't know if this is supposed to have units since it needs to be positions on an x and y-axis and I also don't know how you're supposed to end up with 2 values. Thanks in advance!

 

[anathema]

Thank you for your question. In order to determine the position of the proton, we need to consider the forces acting on it. The electric force exerted by the point charge q is given by Coulomb's law:

F = (k*q*q)/r^2

where k is the Coulomb constant, q is the charge of the proton, and r is the distance between the point charge and the proton. We also have the force of gravity acting on the proton, given by:

Fg = m*g

where m is the mass of the proton and g is the acceleration due to gravity.

In order for the electric force to be exactly opposite to the weight, we need to have F = -Fg. This means that the magnitude of the electric force must be equal to the weight, and the direction must be opposite. So we can set up the following equation:

(k*q*q)/r^2 = -m*g

Substituting in the given values for q, m, and g, we get:

(8.99e9 Nm^2/C^2 * (-0.368e-9 C)^2)/r^2 = -(1.67e-27 kg * 9.8 m/s^2)

Solving for r, we get:

r = 1.12e-3 m

Now, in order to determine the position of the proton, we need to consider the x and y components of this distance. Since the point charge is fixed at the origin, the x-coordinate of the proton must be 0. So we only need to determine the y-coordinate. To do this, we can use the trigonometric relationship:

tan(theta) = y/x

where theta is the angle between the x-axis and the line connecting the proton and the point charge. In this case, the angle is 90 degrees, so we have:

tan(90) = y/1.12e-3

Solving for y, we get:

y = 1.12e-3 m

So the position of the proton is (0, 1.12e-3) on the x-y plane. This means that the proton must be placed 1.12e-3 meters above the origin, in order for the electric force to be exactly opposite to the weight.

I hope this helps! Let me know if you have any further questions.


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[tomcue]

To determine the x and y coordinates of the proton, we need to use the concept of vector addition. The electric force and weight are acting in opposite directions, so their vector sum must be equal to zero. This means that the magnitude of the electric force must be equal to the magnitude of the weight, but in the opposite direction.

Using the given values, we can calculate the magnitude of the electric force as 5.91e-22 N. This means that the magnitude of the weight must also be 5.91e-22 N. We can calculate the weight using the formula m*g, where m is the mass of the proton and g is the acceleration due to gravity. Substituting in the given values, we get 1.64e-27 N.

Now, we can set up a vector addition problem to determine the x and y coordinates of the proton. Let's call the coordinates of the proton as (x,y). The electric force vector can be represented as (-5.91e-22, 0) N, since it is acting in the negative x direction. The weight vector can be represented as (0, -1.64e-27) N, since it is acting in the negative y direction. The vector sum of these two vectors must be equal to zero, so we can write the following equation:

(-5.91e-22, 0) + (0, -1.64e-27) = (0,0)

Solving for x and y, we get x = -5.91e-22 m and y = -1.64e-27 m. Therefore, the proton must be placed at (-5.91e-22, -1.64e-27) on the x- and y-axes, respectively, in order for the electric force acting on it to be exactly opposite to its weight.