If one charge (+5 uC) is placed at origin and a second charge of (+7 uC) at x = 100 cm, then where can a third charge be placed and of what size, so that it has no electrostatic force on it.
I know the magnitude of the force will not matter, however, I'm not sure as to how to determine the distance.
Cheers.
To determine the position and magnitude of the third charge, you need to consider the principle of superposition. The principle states that the net electrostatic force on a charged particle is the vector sum of the forces due to all other charges present.
In this case, you have two charges: +5 uC at the origin and +7 uC at x = 100 cm.
1. Calculate the distance between the two existing charges:
Distance = 100 cm - 0 cm = 100 cm = 1 m
2. Next, find the position where the third charge should be placed. Since we want the net electrostatic force on it to be zero, we need to find a position where the forces due to the other two charges cancel out. In other words, the vectors representing the electrostatic forces due to each charge must have equal magnitudes but opposite directions.
Let's assume the third charge (q3) is placed at a position x meters away from the origin.
The magnitude of the force due to the +5 uC charge at the origin is given by:
F1 = k * |q1 * q3| / r^2 (where k is the electrostatic constant and r is the distance between the charges)
The magnitude of the force due to the +7 uC charge at x = 100 cm is given by:
F2 = k * |q2 * q3| / r^2
Since we want the net force to be zero, F1 = -F2 (forces have opposite directions):
k * |q1 * q3| / r^2 = -k * |q2 * q3| / r^2
3. Substitute the values into the equation and solve for q3:
|q1 * q3| = -|q2 * q3|
|5 * q3| = -|7 * q3|
5 * q3 = -7 * q3 (ignore the negative signs as magnitudes are always positive)
12 * q3 = 0
q3 = 0
Therefore, the magnitude of the third charge (q3) needs to be 0 in order for it to have no electrostatic force on it.
In conclusion, the third charge should be placed at any position along the line between the two existing charges, and its magnitude should be 0 in order for it to experience no net electrostatic force.
To find the location and magnitude of the third charge where it experiences no electrostatic force, you can use Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's denote the distance between the first charge at the origin and the third charge as 'd'. The distance between the second charge at x = 100 cm and the third charge is (100 - d) cm.
The electrostatic force between the first and third charges is given by:
F1-3 = k * |q1| * |q3| / d^2 (Equation 1)
where k is the electrostatic constant and |q1| and |q3| are the magnitudes of the charges.
Similarly, the electrostatic force between the second and third charges is:
F2-3 = k * |q2| * |q3| / (100 - d)^2 (Equation 2)
For the third charge to experience no electrostatic force, both F1-3 and F2-3 must be zero.
Setting the forces to zero and solving the equations, we get:
k * |q1| * |q3| / d^2 = 0
k * |q2| * |q3| / (100 - d)^2 = 0
Since the product of two non-zero numbers divided by zero is undefined, we can conclude that either |q3| = 0 (charge is zero) or d = 0 (distance is zero).
Therefore, if the third charge has zero charge (q3 = 0), it can be placed at any distance, and it will not experience any electrostatic force. Conversely, if the third charge has a non-zero charge, it must be placed at the same location as one of the charges (d = 0) to experience no electrostatic force.
The easy answer is at infinity.
let x be the distance from origin.
F5=F7
kq1 q/x^2= kq2 q/(1-x)^2
note k divides out, q divides out, so the charge on q does not matter.
5E-6 *(1-2x+x^2)= 7E-6*x^2
5-10x+5x^2)-7x^2=0
2x^2+10x-5=0
x=(-10+-sqrt(100+40)/4
x=2.5+-1/4 11.8
x=2.5+-2.95=.45m for one position
x=(10+-sqrt(100+40)/4
x=2.5+-.5*sqrt(140)
=2.5+-.5*11.8