Question

coulomb charge in vector form...... derivation

Answer 1

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Let the position vectors of charges q1 and q2 be r1 and r2 respectively Fig. We denote force on q1 due to q2 by F12 and force on q2 due to q1by F21. The two point charges q1 and q2 have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by r21:



r21 = r2 – r1. In the same way, the vector leading from 2 to 1 is denoted by r12: r12 = r1 – r2 = – r21 .The magnitude of the vectors r21 andr12 is denoted by r21 and r12, respectively (r12 = r21). The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to 1), we define the unit vectors:



Coulomb’s force law between two point charges q1 and q2 located at r1 and r2 is then expressed as



If q­1 and q2 are of the same sign (either both positive and both negative), F21 is along ˆr 21, which denotes repulsion, as it should be for like charges. If q1 and q2 are of opposite signs, F21 is along – ˆr 21(= ˆr 12), which denotes attraction, as expected for unlike charges.



Thus, we do not have to write separate equations for the cases of like and unlike charges. The force F12 on charge q1 due to charge q2, is obtained from force F21, by simply interchanging 1 and 2, i.e.



Thus, Coulomb’s law agrees with the Newton’s third law.



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