Two positive point charges are placed on the x-axis. One, of magnitude 4Q, is placed at the origin. The other, of magnitude Q is placed at x=3 m. Neither charge is able to move. Where on the x-axis in meters can I place a third positive point charge such that the magnitude of the net force on the third charge is zero?
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To determine the position on the x-axis where a third positive point charge can be placed such that the net force on it is zero, we need to find the point where the electric forces from the other two charges balance each other.
Let's denote the charge at the origin (x=0) as Q1 with magnitude 4Q, and the charge at x=3m as Q2 with magnitude Q.
The electric force between two charges is given by Coulomb's law:
F = (k * |Q1 * Q2|) / r^2
Where F is the force, k is the electrostatic constant (9 * 10^9 N m^2/C^2), Q1 and Q2 are the magnitudes of the charges, and r is the distance between them.
Now, we want to find the position on the x-axis where the net force on the third charge is zero, meaning the forces from Q1 and Q2 balance.
Let's assume the third charge, denoted as Q3, is placed at a distance x from the origin. The force due to Q1 on Q3 is:
F1 = (k * |Q1 * Q3|) / x^2
The force due to Q2 on Q3 is:
F2 = (k * |Q2 * Q3|) / (x - 3)^2
For the net force to be zero, F1 and F2 must have equal magnitudes but opposite directions:
F1 = -F2
(k * |Q1 * Q3|) / x^2 = (k * |Q2 * Q3|) / (x - 3)^2
Now, we can cancel out the electrostatic constant k and rearrange the equation:
|Q1 * Q3| / x^2 = |Q2 * Q3| / (x - 3)^2
Since we are looking for the magnitude of the charges, we can ignore the absolute values:
(Q1 * Q3) / x^2 = (Q2 * Q3) / (x - 3)^2
Now, we can substitute the given values Q1 = 4Q and Q2 = Q:
(4Q * Q3) / x^2 = (Q * Q3) / (x - 3)^2
Simplifying further:
(4 * Q * Q3) / x^2 = (1 * Q * Q3) / (x - 3)^2
Cross-multiplying:
(4 * Q * Q3) * (x - 3)^2 = (1 * Q * Q3) * x^2
Expanding and simplifying:
4 * Q * Q3 * (x^2 - 6x + 9) = Q * Q3 * x^2
Canceling common terms Q * Q3 and x^2:
4 * (x^2 - 6x + 9) = x^2
Expanding and simplifying:
4x^2 - 24x + 36 = x^2
Rearranging:
3x^2 - 24x + 36 = 0
This is a quadratic equation. To find the values of x that satisfy this equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 3, b = -24, and c = 36, we can solve for x:
x = (-(-24) ± √((-24)^2 - 4 * 3 * 36)) / (2 * 3)
Simplifying:
x = (24 ± √(576 - 432)) / 6
x = (24 ± √144) / 6
Taking the square root:
x = (24 ± 12) / 6
x = (24 + 12) / 6 or x = (24 - 12) / 6
x = 6 or x = 2
Therefore, you can place the third positive point charge at either x = 6 m or x = 2 m on the x-axis such that the magnitude of the net force on it is zero.