P8. Charges q, and q, lie on the x axis at points x =-5 and x = +5, respectively. (a) How must q and q, be related for the net force on charge +Q, placed at x -+ 5/2, to be zero? (b)Answer the same question if the +Q charge is placed at X=+15/2.
To determine the relationship between the charges q1 and q2 for the net force on charge +Q to be zero, we need to consider the balance of forces acting on charge +Q.
(a) When +Q is placed at x = 5/2:
The net force on +Q can be calculated using Coulomb's Law:
F = k * (q1 * qQ) / r1^2 - k * (q2 * qQ) / r2^2
Here,
- F is the net force on +Q,
- k is the electrostatic constant (k = 9 x 10^9 N m^2 / C^2),
- q1 and q2 are the charges at -5 and +5 respectively,
- qQ is the charge on +Q,
- r1 is the distance between charge q1 and +Q,
- r2 is the distance between charge q2 and +Q.
To make the net force zero, F = 0:
0 = k * (q1 * qQ) / r1^2 - k * (q2 * qQ) / r2^2
Since r1 = r2 = 5/2 - (-5) = 15/2, we can simplify the equation:
0 = k * (q1 * qQ) / (15/2)^2 - k * (q2 * qQ) / (15/2)^2
0 = (q1 * qQ) - (q2 * qQ)
0 = q1 - q2
Therefore, the charges q1 and q2 must be equal in magnitude but opposite in sign for the net force on charge +Q to be zero.
(b) When +Q is placed at x = 15/2:
The same logic applies. The equation becomes:
0 = k * (q1 * qQ) / (15/2)^2 - k * (q2 * qQ) / (15/2)^2
0 = (q1 * qQ) - (q2 * qQ)
0 = q1 - q2
The result is the same, regardless of the position of +Q on the x-axis. The charges q1 and q2 must always be equal in magnitude but opposite in sign for the net force on charge +Q to be zero.