A 2.4 nC charge is at the origin and a -5.6 nC charge is at x = 1.0 cm .
a)At what x-coordinate could you place a proton so that it would experience no net force?
b)Would the net force be zero for an electron placed at the same position?
F = k Q1 Q2/d^2 for each
we want x so F = 0
distances are x and from 2.4 and 1-x from -5.6
for F = 0
2.4/x^2 - 5.6/(1-x)^2 = 0
(NOTE - by inspection you can see that to get zero you must be left of the origin, negative x)
if you locate so F = 0, then the electric field from the two charges is zero there and the force will be zero on any old charge you put there. Does not matter proton or neutron or Uranium nucleus.
Hey, shouldn't it be (x+1)^2
lets say:
q1=2.4
q2=-5.6
q3=proton
q3--x---q1----d-----q2
d=1
Therefore:
F23=K|q2||q3|/(x+1)^2
Also after solving the quadratic I got
x=1.8956...cm
x=-0.395....cm
The question asks for x-cord, x is the distance from the origin. Which x do I choose and what is my final answer? Also why
Remember. The proton is being pushed by the plus charge and pull by the negative. If you put it between the two charges it would be draw right by the positive and pushed right by the negative. If we put it beyond the second charge it's pulled left by a much stronger negative charge and weakly affected right by the positive. The only place they can balance is left of the first charge. Weaker pushing left, stronger pulling right. ANd yes the distance should be x and x+1.
To determine the x-coordinate at which a proton would experience no net force, we need to find the point where the electrical forces due to the two charges cancel each other out. The force between two charges can be calculated using Coulomb's law:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant (9 x 10^9 N m^2/C^2),
q1 and q2 are the charges,
r is the distance between the charges.
a) To find the x-coordinate where the net force on a proton is zero, we need to calculate the forces exerted on it by the two charges.
The force exerted by the 2.4 nC charge at the origin (q1) on the proton (q2) is given by:
F1 = k * (q1 * q2) / r1^2
The force exerted by the -5.6 nC charge at x = 1.0 cm (q3) on the proton (q2) is given by:
F2 = k * (q2 * q3) / r2^2
For there to be no net force on the proton, F1 and F2 must be equal in magnitude. Since the charges have opposite signs, we can compare the magnitudes:
|F1| = |F2|
k * (q1 * q2) / r1^2 = k * (q2 * q3) / r2^2
Now, substitute the values:
(9 x 10^9 N m^2/C^2) * (2.4 x 10^(-9) C * q2) / r1^2 = (9 x 10^9 N m^2/C^2) * (q2 * (-5.6 x 10^(-9) C) / r2^2
Simplifying, we get:
2.4 * q2 / r1^2 = -5.6 * q2 / r2^2
We can cancel out q2 since it appears on both sides:
2.4 / r1^2 = -5.6 / r2^2
Cross multiply:
2.4 * r2^2 = -5.6 * r1^2
Solve for r2^2:
r2^2 = (-5.6 / 2.4) * r1^2
Now, to find the x-coordinate, we can use the distance between the two charges (1.0 cm) and the fact that r1 + r2 = 1.0 cm:
r2 = 1.0 cm - r1
Substitute this value in the equation for r2^2:
(1.0 cm - r1)^2 = (-5.6 / 2.4) * r1^2
Expand:
1.0 cm^2 - 2 * 1.0 cm * r1 + r1^2 = (-5.6 / 2.4) * r1^2
Simplify and rearrange:
(5.6 / 2.4) * r1^2 - 2 * 1.0 cm * r1 - 1.0 cm^2 = 0
Now, you can solve this quadratic equation for r1, and then find the corresponding x-coordinate (which is 1.0 cm - r1).
b) For an electron, which has a negative charge, the net force would not be zero at the same position because the electrical forces would not cancel out. The forces would instead add up, reinforcing each other.
To summarize:
a) The x-coordinate at which a proton would experience no net force can be found by solving the above quadratic equation for r1 and then subtracting it from 1.0 cm to get the x-coordinate.
b) The net force would not be zero for an electron placed at the same position.