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Electric Charges and Fields

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Forces between Multiple Charges

consider a system of three charges q1, q2 and q3, as shown in given figure.The force on one charge, say q1, due to two other charges q2, q3 can, therefore, be obtained by
performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12, F12 is given by,

straight F subscript 12 space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r subscript 12 superscript 2 end fraction straight r with hat on top subscript 12

In the same way, the force on q1 due to q3, denoted by F13, is given by

straight F subscript 13 space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r subscript 13 superscript 2 end fraction space straight r with hat on top subscript 13

which again is the Coulomb force on q1 due to q3, even though other charge q2 is present.

The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n:

straight F subscript 1 space equals space straight F subscript 12 space plus space straight F subscript 13 space plus space..... plus space straight F subscript 1 straight n end subscript space
space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction open square brackets fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r subscript 12 superscript 2 end fraction straight r with hat on top subscript 12 plus space fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r subscript 13 superscript 2 end fraction straight r with hat on top subscript 13.... fraction numerator straight q subscript 1 straight q subscript straight n over denominator straight r subscript 1 straight n end subscript superscript 2 end fraction straight r with hat on top subscript 1 straight n end subscript close square brackets
space equals space fraction numerator straight q subscript 1 over denominator 4 πε subscript 0 end fraction sum from straight i space equals 2 to straight n of fraction numerator straight q subscript straight i over denominator straight r subscript 1 straight i end subscript superscript 2 end fraction straight r with hat on top subscript 1 straight i end subscript

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