Point Charge Problem: Force Exerted by Q1 & Q2 on Q3

[Chandasouk]

Two point charges are placed on the x-axis as follows: one positive charge, Q1= 3.96nC , is located to the right of the origin at x = 0.195m , and a second positive charge, Q2= 5.05nC , is located to the left of the origin at x = -0.295m .

What is the total force (magnitude and direction) exerted by these two charges on a negative point charge, Q3= -5.97nC , that is placed at the origin?

Use 8.85×10−12 for the permittivity of free space. Take positive forces to be along the positive x-axis. Do not use unit vectors.

For my answer, I put 0.000002575N on mastering physics but it said

Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

Here is my work

F1 = (8.988 * 10⁹)[(0.000000004C)(0.000000006)/(0.195m)2] = 0.000005673N

F2=(8.988 * 10⁹)[(0.000000005C)(0.000000006)/(0.295m)2]=0.000003098N

F_x=F1-F2 = 0.000002575N

 

[berkeman]

Two point charges are placed on the x-axis as follows: one positive charge, Q1= 3.96nC , is located to the right of the origin at x = 0.195m , and a second positive charge, Q2= 5.05nC , is located to the left of the origin at x = -0.295m .

What is the total force (magnitude and direction) exerted by these two charges on a negative point charge, Q3= -5.97nC , that is placed at the origin?

Use 8.85×10−12 for the permittivity of free space. Take positive forces to be along the positive x-axis. Do not use unit vectors.

For my answer, I put 0.000002575N on mastering physics but it said

Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

Here is my work

F1 = (8.988 * 10⁹)[(0.000000004C)(0.000000006)/(0.195m)2] = 0.000005673N

F2=(8.988 * 10⁹)[(0.000000005C)(0.000000006)/(0.295m)2]=0.000003098N

F_x=F1-F2 = 0.000002575N

Well, you do appear to be rounding off the charge values. Why are you doing that? I'd suggest 1) use scientific notation in your equations above, and 2) do not round off the charge values. Do you get the right answer now? Everything else looks okay.

 

[Chandasouk]

Thank you, my calculator automatically rounded it for some reason.

 

[berkeman]

Thank you, my calculator automatically rounded it for some reason.

It looks like it's because it wasn't using Scientific Notation maybe, and ran out of room... Glad you figured it out.

 

[paranormal]

Firstly, it is important to note that the question specifically asks for the total force exerted by Q1 and Q2 on Q3, not just the force in the x-direction. Therefore, we need to calculate the total force using vector addition.

To do this, we can use Coulomb's Law which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, we have two point charges (Q1 and Q2) exerting a force on a third point charge (Q3). Therefore, the total force (F) can be calculated as:

F = k * (Q1 * Q3 / r1^2) + k * (Q2 * Q3 / r2^2)

Where k is the Coulomb's constant (8.988 * 10^9 N*m^2/C^2), Q1 and Q2 are the charges of the two point charges (in this case, Q1 = 3.96nC and Q2 = 5.05nC), Q3 is the charge of the point charge at the origin (in this case, Q3 = -5.97nC), and r1 and r2 are the distances between Q1 and Q3, and Q2 and Q3 respectively.

Plugging in the values, we get:

F = (8.988 * 10^9 N*m^2/C^2) * [(3.96 * 10^-9 C * -5.97 * 10^-9 C) / (0.195m)^2] + (8.988 * 10^9 N*m^2/C^2) * [(5.05 * 10^-9 C * -5.97 * 10^-9 C) / (0.295m)^2]

= -0.002183N + -0.001080N

= -0.003263N

Therefore, the total force exerted by Q1 and Q2 on Q3 is -0.003263N. The negative sign indicates that the force is attractive, pulling Q3 towards the origin. This makes sense as Q1 and Q2 are both positive charges, so they will exert a force on the negative charge Q3 towards them.

It is important to note that the magnitude of the force is 0.003