The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)Count said:Homework Statement: A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm.
At what x-coordinate could you place a proton so that it would experience no net force?
Relevant Equations: (3.5/x^2)=(10/(.02+x)^2)
Apparently the answer is not 2.9 cm or 7.4mm. I've looked at similar questions in this thread and solving the exact same way gives me the same answers. I set it up as (3.5/x^2)=(10/(.02+x)^2).
What am I doing wrong?
I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?Orodruin said:The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)
I agree with 2.9cm left of origin. You can rule out the other solution to the quadratic.Count said:I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?
That is your mistake. The statement of the problem asks "At what x-coordinate could you place a proton so that it would experience no net force?" Coordinates to the left of the origin are conventionally negative.Count said:and my mistake may be in not using the negative sign to show the point charge to the left of the origin,
The electric field at a point 5 cm from the origin along the x-axis can be calculated using the formula E = k * |q| / r^2, where k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge. First, calculate the electric field due to the 3.5 nC charge at the origin, then calculate the electric field due to the -10 nC charge at x = 2 cm. Finally, sum the two electric fields to get the total electric field at the given point.
The direction of the electric field at the point 5 cm from the origin along the x-axis can be determined by considering the direction of the electric fields due to each individual charge. Since the 3.5 nC charge at the origin is positive, the electric field points radially outward. On the other hand, the -10 nC charge at x = 2 cm is negative, so the electric field points radially inward. By summing the individual electric fields, you can determine the net direction of the electric field at the given point.
The magnitude of the force on a positive 2 nC charge placed at the point 5 cm from the origin along the x-axis can be calculated using the formula F = q * E, where q is the charge and E is the electric field at that point. First, calculate the electric field at the given point as described in question 1. Then, multiply the charge of the test charge (2 nC) by the magnitude of the electric field to find the force.
The direction of the force on a positive 2 nC charge placed at the point 5 cm from the origin along the x-axis can be determined by the direction of the electric field and the charge of the test charge. Since the test charge is positive, it will experience a force in the same direction as the electric field. Calculate the electric field at the given point and use the charge of the test charge to determine the direction of the force.