Final answer:
Using the work done to bring the charges together and the formula for electric potential, the magnitude of the second charge is calculated to be approximately 3.22 µC.
Step-by-step explanation:
The question involves calculating the magnitude of the charge on a second object using the concept of work and electric potential energy from Physics. We know the work done to bring the charges together is 29 J, and one of the charges is +3.00 µC. The distance between the final positions of the two charges is 2.4 mm (since they are on the same x-coordinate but have a difference of 2.4 mm or 0.0024 meters in their y-coordinates).
To find the magnitude of the charge on the second object, we use the formula for work done in moving a charge in an electric field:
W = VQ
where W is the work done (29 J), V is the electric potential due to the other charge at the point where the +3.00 µC charge is brought, and Q is the charge (which is +3.00 µC in this case).
The electric potential V created by a point charge is given by:
V = k * q / r
where k is Coulomb's constant (8.988 × 109 N*m2/C2), q is the unknown charge we want to find, and r is the distance between the charges.
Combining the equations, we get:
W = (k * q * Q) / r
Solving for q:
q = (W * r) / (k * Q)
Substituting the known values:
q = (29 J * 0.0024 m) / (8.988 × 109 N*m2/C2 * 3.00 µC)
q = (29 * 0.0024) / (8.988 × 109 * 3.00 × 10-6)
q ≈ 3.22 µC
Therefore, the magnitude of the charge on the second object is approximately 3.22 µC.