Which expression gives the force on charge q1 due to charge q2 in the vector form of Coulomb's law?

Which expression gives the force on charge q1 due to charge q2 in the vector form of Coulomb's law?

• A. F12 = (1/(4π ε0)) (q1 q2 / r^2) r_hat_12
• B. F12 = (1/(4π ε0)) (q1 q2 / r^2) r_hat_21
• C. F12 = −(1/(4π ε0)) (q1 q2 / r^2) r_hat_12
• D. F12 = (1/(4π ε0)) (q1 q2 / r^2)

Answer: (A)

The force on q1 due to q2 is directed along the line joining the two charges and has magnitude k |q1 q2| / r^2. In vector form, you express this as F_12 = k q1 q2 / r^2 times a unit vector that points from the source charge q2 to the field point where q1 sits. With r_hat_12 defined to point from q2 toward q1, the expression becomes F_12 = (1/4π ε0) (q1 q2 / r^2) r_hat_12. This directly encodes both the correct magnitude and the correct direction for the force, given that convention. The other options either use the opposite direction for the unit vector, include an extra minus sign, or omit the directional factor, which would fail to capture the vector nature of the force.

The force on q1 due to q2 is directed along the line joining the two charges and has magnitude k |q1 q2| / r^2. In vector form, you express this as F_12 = k q1 q2 / r^2 times a unit vector that points from the source charge q2 to the field point where q1 sits. With r_hat_12 defined to point from q2 toward q1, the expression becomes F_12 = (1/4π ε0) (q1 q2 / r^2) r_hat_12. This directly encodes both the correct magnitude and the correct direction for the force, given that convention. The other options either use the opposite direction for the unit vector, include an extra minus sign, or omit the directional factor, which would fail to capture the vector nature of the force.