What Is Vector Trap?

Vector form of Coulomb's law?

• From the formula of unit vector we have r(unit vector)= r(vector)/r(magnitude) therefore r(vector)= r(unit vector).r(magnitude).For Coulomb's law the formula is F=K.q1q2/ r^2 and its vector form becomes F(vector)= K.q1q2.r(unit vector)/r(magnitude), according to the above explained unit vector formula shouldn't the vector form of coulomb's be F=K.q1q2/r(magnitude).r(unit vector)? How can the r(unit vector) be in numerator? Need help!

• Answer:

  You can't just replace the magnitude of "r" with that vector eq. You have to think of kq1q2/r^2 as the magnitude of the force. "r" in this formula is a magnitude. Then to write the force as a vector you simply affix the unit vector to the magnitude. So the eq is; F(vector) = (kq1q2/r^2)r(unit vector) Now it is permissible to re write the unit vector as ;r(unit vector) = r(vector)/r(magnitude) to get; F(vector) = (kq1q2/r^3)r(vector) Also, if you were wondering, you can't write a scalar as the ratio of two vectors as; r(mag) = r(vector)/r(unit vector) ,because division of vectors is not a defined operation.