Two charges, separated by a distance of 1.3 m, attract each other with a force of 0.078
N. If one charge is +3.7 μC, what is the other charge?
F = (kq1q2)/r^2
0.078 N = (9 * 10^9 N*m^2/C^2)(3.7 * 10^-6 C)(q2)/(1.3 m)^2
q2 = 0.078 N * (1.3 m)^2/((9 * 10^9 N*m^2/C^2)(3.7 * 10^-6 C))
q2 = _______ C
Also note that q2 will be negative because the force is attraction.
To solve this problem, we will use Coulomb's law, which states that the force between two charges is given by the equation:
F = k * (q1 * q2) / r^2
where F is the force between the charges, k is the electrostatic constant (9 × 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
We are given the force (F = 0.078 N), the distance (r = 1.3 m), and one of the charges (q1 = +3.7 μC = 3.7 × 10^-6 C). We need to find the other charge (q2).
Rearranging the equation, we get:
q2 = (F * r^2) / (k * q1)
Substituting the known values, we can calculate the other charge:
q2 = (0.078 * (1.3^2)) / (9 × 10^9 * 3.7 × 10^-6)
Now, let's calculate it:
q2 = 0.078 * 1.69 / (9 × 10^9 * 3.7 × 10^-6)
To simplify, perform the calculations inside the parentheses first:
q2 = 0.13242 / (9 × 10^9 * 3.7 × 10^-6)
Now, let's simplify the calculations inside the parentheses:
q2 = 0.13242 / (3.33 × 10^4)
Dividing these two numbers gives:
q2 = 3.97 × 10^-6 C
Therefore, the other charge is +3.97 μC.