Coulomb’s Law describes the force between two (and only two!) charged objects. S
ID: 249734 • Letter: C
Question
Coulomb’s Law describes the force between two (and only two!) charged objects. Suppose Object 1 has a charge of +1.75 C and Object 2 has a charge of +12.9 C Object 2 is located a distance of 0.03340 m to the right of Object 1 (Yes — please break down and draw a quick sketch!). What is the magnitude and direction of the electrical force on Object 2 due to Object 1? to the left to the right Submit Answer Tries 0/10 What is the magnitude and direction of the electrical force on Object 1 due to Object 2? to the left to the right Submit Answer Tries 0/10 You now move Object 2 such that it is located a distance of 0.06680 m to the right of Object 1 (twice the distance as before). What is the magnitude and direction of the electrical force on Object 2 due to Object 1? to the left to the right Submit Answer Tries 0/10 The new force on Object 1 due to Object 2 (at a separation of 0.06680 m) is ____ times the original force on Object 1 due to Object 2 (at a separation of 0.03340 m)? (Provide a numerical answer to fill in the blank.)
Explanation / Answer
The magnitude of the electric force on object 1 due to object 2 = the magnitude of the electric force on object 2 due
to object 1 = kq1*q2/r^2 = 8.99*10^9*1.75*10^(-6)*12.9*10^(-6)/(0.0334)^2 = 181.92 N
The direction of the electric force on object 1 is away from the object 2 and vice versa.
now the distance is twice so
The magnitude of the electric force on object 1 due to object 2 = the magnitude of the electric force on object 2 due
to object 1 = kq1*q2/r^2 = 8.99*10^9*1.75*10^(-6)*12.9*10^(-6)/(0.0668)^2 = 45.48 N
The new force on Object 1 due to Object 2 is 1/4 times the original force on Object
1 due to Object 2